Diffraction and interference

Class 12 Physics, unit 18, book pages 70–85

When waves overlap they add (superposition), which makes some places stronger and others weaker: this is interference. Waves also spread around edges and through narrow gaps: this is diffraction. The unit uses these ideas to explain two-source patterns with water, sound, light and microwaves, Young's double-slit experiment (), and the diffraction grating () used to measure wavelengths. It ends with the electron double-slit experiment, which shows that matter also behaves as a wave, and two interpretations of what this means.

Learning outcomes (6)
  • Explain experiments that demonstrate two-source interference using water waves in a ripple tank, sound, light and microwaves.
  • Describe the conditions required if two-source interference fringes are to be observed.
  • Use for double-slit interference using light to solve problems.
  • Use to solve problems.
  • Describe the use of a diffraction grating to determine the wavelength of light (the structure and use of the spectrometer are not included).
  • In the context of the electron double-slit experiment, explain two of the many interpretations of quantum mechanics: (i) the Copenhagen interpretation and (ii) the many-worlds interpretation.

Superposition, interference and diffraction

Water ripples, sound, light and microwaves are all waves. When two waves meet, they do not bounce off each other like balls. They pass through each other, and while they overlap their displacements simply add. This one idea, the principle of superposition, explains both topics of this unit: interference and diffraction.

Principle of superposition
When two or more waves overlap at a point, the resultant displacement is the sum of the displacements of the individual waves at that point.
Interference
The effect produced when two or more waves overlap and combine to give a resultant wave of larger or smaller amplitude.
Diffraction
The spreading (bending) of waves around the edges of an obstacle or through a narrow opening into the region of geometrical shadow.
superposition

Here and are the displacements that each wave alone would produce at a point, and is the resultant displacement.

Constructive and destructive interference

If the waves arrive in phase (crest on crest, trough on trough), they add to give a wave of larger amplitude. This is constructive interference: brighter light, louder sound, bigger ripples.

Constructive interference: waves in phase
−2
−1
1
2
O
Two equal waves in phase (dashed, on top of each other) add to a wave of double amplitude (solid).

If the waves arrive in anti-phase (phase difference rad or , crest on trough), they cancel. This is destructive interference: darkness, silence or calm water.

Destructive interference: waves in anti-phase
−1
1
O
Two equal waves in anti-phase cancel completely at every instant.

Diffraction depends on the size of the gap

When a wave passes through a gap or past an edge, it spreads into the region behind the obstacle. Every point on a wavefront acts as a source of new wavelets (Huygens' principle), and at the edges nothing cancels the sideways-moving wavelets. The spreading is large when the gap width is about the same as the wavelength , and small when the gap is much wider than .

Plane waves meeting a wide gap (left) and a narrow gap (right). The narrower the gap compared with , the more the waves spread.
Interference and diffraction compared: a common short question.
FeatureInterference (two slits)Diffraction (single slit)
What overlapswaves from two separate coherent sourceswavelets from different parts of the same wavefront
Width of fringesall fringes equally widecentral band twice as wide as the others
Brightnessbright fringes nearly equally brightbrightness falls quickly away from the centre

18.1Two-source interference and conditions for fringesOften examined

Two-source interference (also called double-slit interference) happens when waves from two coherent sources overlap. At some points the waves always reinforce each other and at other points they always cancel, so a fixed pattern of strong and weak regions, called fringes, appears. It can be shown with water waves in a ripple tank, sound waves, light waves and microwaves.

Coherent sources
Two sources that emit waves of the same frequency (and so the same wavelength) with a constant phase difference between them.
Monochromatic light
Light of a single wavelength, that is, of a single colour.
Path difference
The difference between the distances travelled by two waves from their sources to the same point, . SI unit: metre.
Phase difference
The difference in the phases of two waves at a point, measured in radians or degrees.

Path difference and phase difference

Two coherent sources that start in phase can still produce waves that arrive out of phase, because the waves travel different distances. The book says this simply: the phase difference is produced by the path difference.

Conditions for constructive and destructive interference

Here
Constructive interferenceDestructive interference
Waves arrivein phase (crest on crest)in anti-phase (crest on trough)
Path difference
Phase difference
Resultlarge amplitude: bright fringe, loud sound, large ripplessmall or zero amplitude: dark fringe, quiet sound, calm water

Conditions for observing interference fringes

A steady (sustained) pattern that we can see or measure needs the conditions below. The first four are the book's list; the last two make the fringes clear and easy to see.

  1. Coherent sources. The two sources must have the same frequency and a constant phase difference. If the phase difference keeps changing, the bright and dark fringes swap places millions of times per second and the eye sees only even illumination. For light, two coherent sources are made from one source by splitting its wavefront (Young's double slit), or a laser is used.
  2. Narrow slits. Each slit must be narrow (its width comparable to the wavelength), so that light spreads out from it and the beams from the two slits overlap.
  3. Monochromatic waves. The waves should have a single wavelength. Different wavelengths give fringes of different spacing, which overlap and blur the pattern.
  4. Stable environment. The set-up must be free of vibrations and air currents, which would change the path difference and make the fringes move.
  5. Small slit separation and a distant screen. The fringe spacing is large enough to see only if the slit separation is small (a fraction of a millimetre for light) and the screen distance is large (about ).
  6. Nearly equal amplitudes. When the two waves have equal amplitudes, destructive interference gives complete darkness, so the fringes have good contrast.

18.1.1Interference of water waves

A ripple tank is a shallow, glass-bottomed tray of water with a lamp above it. The lamp throws a shadow of the ripples onto white paper below the tank: each crest acts like a converging lens and makes a bright line, and each trough makes a dark line. So the moving water pattern can be seen clearly on the paper.

Ripple tank for two-source interference. One motor drives both dippers, so and are coherent.

To make two coherent sources, two small dippers are fixed to the same vibrating bar. Because one motor drives both, they vibrate with the same frequency and stay in phase. Each dipper sends out circular waves, and where the two sets of waves overlap they interfere.

Top view of the interference pattern of two coherent point sources. The antinodal and nodal lines spread out from between the sources like the fingers of a hand.
  • Antinodal lines (A): crest meets crest and trough meets trough, so the water moves up and down strongly and the paper shows sharp bright and dark bands. Path difference .
  • Nodal lines (N): crest meets trough, so the water stays almost still and the paper shows an even grey. Path difference .
  • The central line , the same distance from both sources, is always antinodal, because its path difference is zero.

What changes the pattern

  • Higher frequency (shorter wavelength): more nodal and antinodal lines, packed closer together.
  • Larger source separation : more lines, closer together.
  • Only one dipper: plain circular waves, with no nodal lines, so no interference.
  • Dippers out of step by half a cycle: the whole pattern shifts, and the central line becomes nodal.

18.1.2Interference of sound waves

Two loudspeakers connected to the same signal generator and amplifier give out sound of the same frequency, in phase, so they are coherent sources. Sound is a longitudinal wave made of compressions (high pressure) and rarefactions (low pressure).

Two loudspeakers driven by one signal generator. A microphone moved along a line in front of them finds alternate loud and quiet positions.

Where a compression from one speaker arrives together with a compression from the other (or a rarefaction with a rarefaction), the waves interfere constructively and the sound is loud. Where a compression arrives with a rarefaction, they interfere destructively and the sound is quiet. In the book's Fig. 18.4, points , , , and are loud and points , , and are quiet.

A person walking slowly across the room in front of the speakers hears the sound grow louder and softer. A microphone connected to a meter or an oscilloscope (CRO) shows the same thing as a rise and fall of the signal.

Beats: interference in time

If two sources have slightly different frequencies (for example tuning forks of and ), they are not coherent. At one place the two waves drift in and out of step. When they are in phase the sound is loud; a short time later they are in anti-phase and the sound is soft. This regular rise and fall of loudness is called beats. The number of beats heard per second equals the difference between the two frequencies, so forks of and give beats per second.

Beats from and waves
0.5
1
−2
−1
1
2
O
The resultant of two waves of and . Its loudness rises and falls times per second.

18.1.3Interference of light: Young's double-slit experimentOften examined

Light waves are very short (about to ), so light interference is much harder to see than ripple-tank interference. In 1801 the English physicist and physician Thomas Young found a way to do it. At that time scientists disagreed about light: Huygens described it as a wave, while Newton's followers explained it as a stream of particles. Young's experiment showed clearly that light behaves as a wave.

Apparatus

Young's double-slit experiment: source, single slit , double slit (separation ) and a screen at distance . Bright and dark fringes form where the waves from and overlap.
  • Source: monochromatic light (a sodium lamp, or a laser in modern versions). Young used sunlight.
  • Single slit : a narrow slit that lets through light from one small part of the source. Light spreading from reaches both slits of the double slit from the same wavefront.
  • Double slit , : two narrow parallel slits, a small distance apart (about to ), equidistant from . Because their light comes from the same wavefront, and act as two coherent sources in phase.
  • Screen: placed a distance (about ) from the double slit.

Working

Light from spreads (diffracts) and falls on both slits. Each slit diffracts the light again and sends out semicircular wavefronts. The two sets of waves overlap in the space in front of the slits and interfere. On the screen we see alternate bright and dark bands called fringes, parallel to the slits (book Fig. 18.6).

  • At the centre O the paths and are equal, so the waves arrive in phase: the central fringe is bright (the central maximum, ).
  • Where the path difference is the waves again arrive in phase: bright fringes.
  • Where the path difference is the waves arrive in anti-phase: dark fringes.
  • The fringes are equally spaced, and the bright fringes are all nearly equally bright.

Using white light

Young used sunlight, which is a mixture of wavelengths. Each colour forms its own set of fringes with its own spacing (spacing is proportional to ). At the centre every colour has zero path difference, so the central fringe is white. On each side the fringes are coloured, with violet (short ) on the inside and red (long ) on the outside. After a few fringes the colours overlap and the pattern fades to even white. For a clear pattern with many fringes, monochromatic light is used.

The book's Fig. 18.6 (c) shows that the double-slit pattern of water waves looks almost the same as that of light. This close match is why Young's experiment is taken as evidence that light is a wave.

18.1.3Path difference, fringe positions and fringe spacingOften examined

We now find exactly where the bright and dark fringes form. The symbols are: , the separation of the slits; , the distance from the slits to the screen; , the distance of a point P on the screen from the centre O; and , the angle between the central line MO and the direction MP.

Derivation Path difference at a point on the screen

Aim. Show that the path difference between the waves reaching P from and is .

Assumptions.
  • The screen distance is very large compared with the slit separation (), so the rays and are almost parallel.
  • and are coherent sources in phase.
Geometry of Young's experiment (not to scale: in reality is thousands of times larger than ). BC is the extra distance travelled by the wave from .
  1. Draw AC from (point A) perpendicular to the ray . The screen is very far away, so P is practically the same distance from A and from C.
  2. The wave from (point B) therefore travels an extra distance BC. This is the path difference.
  3. The angle BAC equals , because AC is perpendicular to the ray and AB is perpendicular to the central line (angles between two pairs of perpendicular lines are equal). In the right-angled triangle ABC the hypotenuse is .
  4. Rearrange to get BC.
Result

The kind of interference at P depends only on the slit separation and the direction of P. The path difference is zero at the centre and grows steadily as we move away from it.

Using the conditions for interference with path difference :

bright fringes (18.2)
dark fringes (18.3)

The integer is the order of the fringe. is the central maximum; are the two first-order bright fringes, one on each side of O, and so on. In the book's numbering the first dark fringe has and is called the first-order minimum.

Derivation Positions of bright and dark fringes

Aim. Find the distance of the th bright fringe and the th dark fringe from the centre O.

Assumptions.
  • The angle is small (a few degrees), because is much smaller than .
  1. Start from the condition for the th bright fringe.
  2. The angle is small, so .
  3. In the right-angled triangle formed by MO, OP and MP, .
  4. Solve for .
  5. Repeat with the dark-fringe condition .
Result

The central maximum () is always at . The bright fringes lie at and each dark fringe lies exactly halfway between two bright ones.

Derivation Fringe spacing

Aim. Show that the distance between two neighbouring bright (or dark) fringes is .

  1. The fringe spacing is the distance between two consecutive bright fringes, the th and the th.
  2. Use equation (18.4) for each fringe.
  3. The terms containing cancel.
  4. For dark fringes use (18.5). Again the terms cancel and the spacing is the same.
Result

The spacing does not depend on , so the fringes are evenly spaced, and bright and dark fringes have the same spacing. Measuring , and gives the wavelength of light: .

Intensity across the screen
O
Bright fringes are equally spaced, apart, with dark fringes exactly halfway between them.
Fringe spacing (fringe width)
The distance between the centres of two consecutive bright fringes, or of two consecutive dark fringes: . SI unit: metre.

Factors that change the fringe spacing

ChangeEffect on Reason
increase the slit separation decreases (fringes closer)
increase the screen distance increases (fringes wider)
use red light instead of blueincreases
put the apparatus in water (refractive index )decreasesthe wavelength in water is
use white lightcentral fringe white, coloured fringes beside iteach colour has its own
Example 18.1 (corrected)
The distance between the central maximum and the 1st minimum is . If light of wavelength is used, find the separation between the slits. The distance between the slits and the screen is .
Given
  • central maximum to first dark fringe
To find
  • slit separation
  1. The first dark fringe has in equation (18.5), so it is half a fringe spacing from the centre.
  2. Rearrange for .
  3. Substitute the values in SI units.
Assignment 18.1
Light from a He-Ne laser passes through two slits separated by . The third bright line on a screen is formed at an angle of relative to the incident beam. What is the wavelength of the light?
Given
  • third bright line
To find
  • wavelength
  1. Bright lines satisfy the double-slit condition. The angle is given, so no small-angle approximation is needed.
  2. Substitute.
  3. Evaluate.

18.1.4Interference of microwaves

Microwaves are electromagnetic waves with wavelengths of a few centimetres (3\,\mathrm{cm} at a frequency of 10\,\mathrm{GHz}). Because the wavelength is large, the slits can be gaps a few centimetres wide between metal plates, and the interference pattern is spread over a large region, so it is easy to measure.

Microwave two-source interference. The detector is moved along the dashed arc; its meter shows maxima (filled) and minima (open).
  • Transmitter: sends a beam of microwaves of one frequency towards the metal plates.
  • Metal plates with two slits: metal reflects microwaves, so waves pass only through the two gaps. Both gaps receive the same wavefront, so they act as two coherent sources in phase.
  • Detector (receiver probe) connected to a meter: moved along an arc or a line in front of the slits to measure the intensity.

As the detector moves, the meter reading rises and falls: it is high where the waves from the two gaps interfere constructively and very low (almost zero) where they interfere destructively. The pattern can also be shown on a recording device connected to the detector.

Finding the wavelength of microwaves

  1. Place the detector at a maximum and measure its distances from the two gaps, S_1P and S_2P. Count the order m of this maximum from the centre.
  2. At the mth maximum the path difference is m whole wavelengths.
    S_2P-S_1P=m\lambda
  3. So the wavelength is the path difference divided by m. Alternatively, measure the angle \theta of the maximum and use d\sin\theta=m\lambda.
    \lambda=\dfrac{S_2P-S_1P}{m}

18.2Diffraction grating and grating elementOften examined

A diffraction grating is a plate of glass or plastic with a very large number of equally spaced, parallel slits (lines) ruled or etched on it, for example 600 lines per millimetre. The clear gaps between the ruled lines act as slits. When light falls on the grating, each slit becomes a source of secondary waves, and the waves from all the slits interfere.

Diffraction grating
An optical device with a large number of closely and equally spaced parallel slits (lines or grooves). It diffracts light into sharp bright maxima at definite angles.
Grating element (grating constant) d
The distance between the centres of two adjacent slits of a grating. It equals the width of one slit plus the width of one opaque line, d=a+b. SI unit: metre.

If N lines are ruled on each unit length of the grating, the grating element is

d=\dfrac{\text{unit length of grating}}{\text{total number of lines ruled on it}}=\dfrac{1}{N}
grating element
Grating element from lines per unit length
Find the grating element of (a) a grating with 600 lines per millimetre, (b) a grating with 5000 lines per centimetre.
  1. (a) One millimetre holds 600 lines, so the distance from one line to the next is 1\,\mathrm{mm} divided by 600.
    d=\dfrac{1\times10^{-3}\,\mathrm{m}}{600}=1.67\times10^{-6}\,\mathrm{m}=1.67\,\mu\mathrm{m}
  2. (b) One centimetre holds 5000 lines.
    d=\dfrac{1\times10^{-2}\,\mathrm{m}}{5000}=2.0\times10^{-6}\,\mathrm{m}=2.0\,\mu\mathrm{m}
d=1.67\,\mu\mathrm{m}\ \text{and}\ 2.0\,\mu\mathrm{m}

In a transmission grating light passes through the clear gaps. In a reflection grating light reflects from many fine, parallel grooves; the surface of a CD or DVD behaves like one.

The number of lines per unit length decides how far apart the orders are. The total number of lines covered by the beam decides how sharp the maxima are, and so how well the grating separates two wavelengths that are close together (its resolving power).

18.2.1Diffraction of light through a gratingOften examined

When monochromatic light falls normally (at right angles) on a grating, the light that comes through is not spread evenly. It leaves only in a few sharp, bright beams at definite angles. In between, the waves from the many slits cancel almost completely. The principle of the grating is therefore interference of the waves diffracted by all the slits.

Derivation The grating equation d\sin\theta=m\lambda

Aim. Find the directions \theta in which a grating gives bright maxima.

Assumptions.
  • Plane monochromatic waves of wavelength \lambda fall normally on the grating, so all slits are in phase.
  • The light is observed far away or brought to a focus by a lens, so the rays leaving the slits at one angle are parallel.
Rays leaving neighbouring slits at angle \theta. Each slit's ray is d\sin\theta behind the one above it (red segments).
  1. Consider the rays that leave every slit at the same angle \theta to the normal. Draw the line through the top slit perpendicular to these rays (dashed): it is a wavefront. From this line onwards, a lens brings all the rays together at one point P, and all of them travel equal optical paths to P.
  2. Compare two neighbouring slits, a distance d apart. The ray from the lower slit travels an extra distance before it reaches the dashed line. This distance is the side opposite \theta in a right-angled triangle with hypotenuse d.
    \Delta x=d\sin\theta
  3. The slits are equally spaced, so the next slit is 2d\sin\theta behind the first, the one after that 3d\sin\theta, and so on.
    0,\ d\sin\theta,\ 2d\sin\theta,\ 3d\sin\theta,\ \dots
  4. If the path difference between neighbours is a whole number of wavelengths, every wave reaches P in phase with every other wave. They all reinforce, giving a bright principal maximum.
    d\sin\theta=m\lambda,\qquad m=0,1,2,\dots
  5. At other angles the waves from the many slits arrive with all sorts of phases and cancel almost completely, so it is dark between the maxima.
Resultd\sin\theta=m\lambda

The angles of the bright lines depend only on d and \lambda. If d is known and \theta is measured for a known order m, the wavelength is \lambda=d\sin\theta/m.

\lambda=\dfrac{d\sin\theta}{m}
wavelength from a grating
Monochromatic light through a grating: a sharp central maximum (m=0) and a few sharp orders on each side at angles \theta_1, \theta_2.

Determining the wavelength of light with a grating

  1. Let a parallel beam of monochromatic light (for example from a sodium lamp) fall normally on a grating with a known number of lines per unit length N, so that d=1/N.
  2. Find the central maximum (\theta=0). Then find the first-order maxima on both sides and measure the angle between them; half of this angle is \theta_1.
  3. Repeat for the second order if it can be seen.
  4. Calculate \lambda=d\sin\theta/m for each order and take the average.

Why grating maxima are sharper than double-slit fringes

Double slit and grating with the same slit spacing
Both have maxima at d\sin\theta=m\lambda (at angles \theta_1, \theta_2, ... for orders 1, 2, ...), but the grating maxima are much narrower and the regions between them are darker (book Fig. 18.10). Each curve is scaled to the same peak height; the grating peaks are really far brighter.

A double slit and a grating with the same slit spacing d give maxima at the same angles. But with many slits, a very small change of angle away from d\sin\theta=m\lambda puts the waves from distant slits out of step, and they cancel. So the maxima become very narrow and bright, and the regions between them become darker, as the number of slits increases.

FeatureDouble slitDiffraction grating
Number of slitstwothousands
Bright maximabroad, brightness changes graduallyvery sharp, narrow and bright
Between the maximadark fringes as wide as the bright onesalmost completely dark
Separation of maximasmall angles, many fringes (d large)large angles, few orders (d small)
With white lighta few coloured fringes, then whiteclear, separate spectra in each order

White light and spectra

With white light the central maximum (m=0) is white, because \theta=0 for every wavelength. In every other order each colour appears at its own angle, \sin\theta=m\lambda/d, so red (longest wavelength) is diffracted most and violet least. Each order is a spectrum with violet on the inside and red on the outside. Higher orders are more spread out, and the second- and third-order spectra overlap.

White light through a grating with 600 lines per mm: white central maximum and a first-order spectrum on each side, from violet (V) to red (R).
Example 18.2
Diffracted light is observed at an angle of 30^\circ in the 2nd order (m=2) with a grating constant of 2.0\,\mu\mathrm{m}. Find the wavelength of the diffracted light.
Given
  • \theta = 30^\circ
  • m = 2
  • d = 2.0\,\mu\mathrm{m}=2.0\times10^{-6}\,\mathrm{m}
To find
  • wavelength \lambda
  1. Use the grating equation and solve for \lambda.
    m\lambda=d\sin\theta\;\Rightarrow\;\lambda=\dfrac{d\sin\theta}{m}
  2. Substitute; \sin30^\circ=0.5.
    \lambda=\dfrac{(2.0\times10^{-6})(0.5)}{2}=5.0\times10^{-7}\,\mathrm{m}
\lambda=0.50\,\mu\mathrm{m}=500\,\mathrm{nm}

How the pattern changes

  • More lines per unit length (smaller d): each order moves to a larger angle, so the maxima spread farther apart; fewer orders are possible, because \sin heta=m\lambda/d cannot be more than 1; the maxima become sharper.
  • More slits illuminated (same d): the maxima stay at the same angles but become narrower and brighter, with darker regions between them.
  • Narrower slits (same d): each slit spreads its light more widely, so the higher orders are brighter. With wider slits most of the light stays near the centre and the higher orders become faint.
  • Longer wavelength: every order appears at a larger angle.

18.2.2Applications of diffraction gratings

  1. Wavelength measurement: a diffraction grating is used to determine the wavelength of light accurately.
  2. Atomic spectra analysis: every element emits its own set of wavelengths (its line spectrum). A grating separates them, so elements can be identified in a laboratory sample or in a distant star.
  3. Biomedical imaging: gratings select particular wavelengths, for example to examine biopsy samples for signs of disease.
  4. Spectroscopy: gratings disperse light into its wavelengths for detailed analysis in spectrometers used in chemistry, astronomy and environmental monitoring.

Gratings in nature and daily life

Australian opal contains tiny silica spheres packed in regular rows, and butterfly wings carry regular rows of microscopic scales and ridges. These act like reflection gratings: they send different colours in different directions, so the colours change as the opal or the wing is tilted (book Fig. 18.11). Such colours are called structural colours.

A CD or DVD stores data along a spiral track. On a CD neighbouring turns of the track are 1.6\,\mu\mathrm{m} apart, close to the wavelength of visible light, so the surface acts as a reflection grating and shows rainbow colours. These colours are only a side effect; the track exists to store data.

X-ray diffraction by crystals

X-rays have wavelengths of about 10^{-10}\,\mathrm{m}, similar to the spacing of atoms in a crystal. A crystal therefore acts as a three-dimensional diffraction grating for X-rays. This is the basis of X-ray crystallography, mentioned in the book's introduction.

Diffraction at a single slit

When light passes through one narrow slit of width a, the screen shows a broad, bright central band with much weaker bands on each side. By Huygens' principle every point in the slit sends out wavelets, and the wavelets from different parts of the same slit interfere.

Derivation Condition for the dark bands

Aim. Find the angles \theta at which a single slit gives darkness.

Assumptions.
  • Plane monochromatic waves fall normally on the slit, and the screen is far away.
Rays from the top edge A, the middle M and the bottom edge B of a slit of width a.
  1. Divide the slit into two equal halves, A to M and M to B. Pair each point in the top half with the point a distance a/2 below it (A with M, and so on down to M with B).
  2. In a direction \theta, the two wavelets of each pair have a path difference of (a/2)\sin\theta.
    \Delta x=\dfrac{a}{2}\sin\theta
  3. If this is half a wavelength, the two wavelets of every pair cancel, and the whole slit gives darkness in that direction.
    \dfrac{a}{2}\sin\theta=\dfrac{\lambda}{2}\;\Rightarrow\;a\sin\theta=\lambda
  4. Dividing the slit into 4, 6, 8, ... equal parts in the same way gives the other dark bands.
    a\sin\theta=m\lambda,\qquad m=\pm1,\pm2,\pm3,\dots
Resulta\sin\theta=m\lambda\quad(\text{dark bands, } m\neq0)

A narrower slit (smaller a) gives larger angles, so the pattern spreads out more. m=0 is not a dark band: the centre is the brightest point of the pattern.

Intensity pattern of a single slit
The central maximum is twice as wide as the others and much brighter. Dark bands lie at a\sin\theta=\pm\lambda,\ \pm2\lambda,\dots, that is at \theta\approx\pm\lambda/a,\ \pm2\lambda/a,\dots for small angles.
  • The central maximum is twice as wide as the other bright bands and much brighter.
  • Its edges are at \sin\theta=\pm\lambda/a. On a screen at distance L its full width is about 2\lambda L/a.
  • Increasing the slit width a makes all the bands narrower and closer together. When a is much larger than \lambda the pattern shrinks to a sharp image of the slit.
  • A first dark band exists only if \lambda\le a, because \sin\theta cannot exceed 1.
Assignment 18.2
Visible light of wavelength 550\,\mathrm{nm} falls on a single slit and produces its second diffraction minimum at an angle of 45^\circ relative to the incident direction of the light. What is the width of the slit?
Given
  • \lambda = 550\,\mathrm{nm}=5.50\times10^{-7}\,\mathrm{m}
  • m = 2 second minimum
  • \theta = 45^\circ
To find
  • slit width a
  1. Dark bands of a single slit satisfy a\sin\theta=m\lambda. Solve for a.
    a=\dfrac{m\lambda}{\sin\theta}
  2. Substitute; \sin45^\circ=0.7071.
    a=\dfrac{2\,(5.50\times10^{-7})}{0.7071}=1.56\times10^{-6}\,\mathrm{m}
a=1.56\times10^{-6}\,\mathrm{m}=1.56\,\mu\mathrm{m}

18.3Electron double-slit experiment

The electron double-slit experiment shows that electrons, which we normally picture as tiny particles, also behave like waves. It is a direct demonstration of wave-particle duality and of the nature of reality at the quantum level.

Electron double-slit experiment. Each electron lands as one dot; many dots together form interference fringes.
  • Electron gun: a heated filament emits electrons, which are accelerated into a narrow beam.
  • Double slit: two very narrow parallel slits. They must be far narrower than slits for light, because the wavelength of an electron is tiny.
  • Screen: a fluorescent screen or detector that shows where each electron lands.

What we expect and what we see

If electrons were classical particles, each would go through one slit or the other, and the screen would show just two bands, one behind each slit. Instead the screen shows an interference pattern of many bright and dark bands, just like light in Young's experiment.

Even when electrons are sent one at a time, each electron arrives as a single dot at one point: particle-like behaviour. But after thousands of electrons have arrived, the dots build up the interference pattern: wave-like behaviour. So the motion of each electron is described by a wave that passes through both slits and interferes with itself.

Why the experiment matters

  • Quantum mechanics: it laid the foundation for quantum mechanics, which describes the behaviour of particles at the atomic and subatomic level.
  • Limits of classical physics: classical physics cannot explain the pattern. Particles at the quantum level do not follow definite paths.
  • Philosophical implications: it raises questions about the nature of reality, about observation, and about the role of the observer. If a detector is placed at the slits to find which slit each electron passes through, the interference pattern disappears.

Physicists agree on what is observed but not on what it means. The two best-known interpretations are the Copenhagen interpretation and the many-worlds interpretation.

18.3 (i)Copenhagen interpretation

The Copenhagen interpretation is a widely held interpretation of quantum mechanics, developed by Niels Bohr and Werner Heisenberg. It states that a quantum system, such as an electron before it is measured, does not have a definite position or momentum. Instead it exists in a superposition of all its possible states at the same time, described by a wave function.

How it explains the electron double-slit experiment

  • Before detection the electron is in a superposition of states, so it passes through both slits at once (wave-like behaviour).
  • The act of measurement, detecting the electron on the screen, forces the wave function to collapse into one definite position. That is why we see a single electron hit at one point on the screen (particle-like behaviour).
  • The interference pattern is a pattern of probability. Bright fringes are where the probability of detecting the electron is high; dark fringes are where it is low.

Criticisms

  • The 'collapse of the wave function' on measurement seems like an arbitrary break from the smooth, deterministic laws of physics, and it is not clear what exactly counts as a measurement.
  • It does not explain what happens to the 'other possibilities' in the superposition after the measurement.

The Copenhagen interpretation remains widely held and influential, but its implications and limits are still debated among physicists and philosophers. Its key ideas are indeterminism (only probabilities can be predicted), complementarity (the wave and particle pictures are both needed, but not at the same time) and the central role of measurement.

18.3 (ii)Many-worlds interpretation

The many-worlds interpretation was proposed by Hugh Everett III in 1957. It says that the wave function never collapses. Instead, every quantum interaction with several possible outcomes splits the universe into several realities (worlds), one for each outcome. Each branch of reality carries forward one possibility from the superposition.

How it explains the electron double-slit experiment

  • The electron starts in a superposition, going through both slits.
  • When it reaches the screen the universe branches. In each world the electron interacts with the detector screen and leaves its mark at one particular place (particle-like behaviour).
  • The interference pattern reflects the combined effect of electrons hitting the screen in all the parallel universes. We experience only one world, the one in which the electron struck our screen at a particular point, but the pattern shows that the other possibilities exist too.

Criticisms

  • The other universes cannot be observed, so the idea is very difficult to test by experiment.
  • It does not give a clear explanation of the change from a superposition to the one definite result that we observe.
The two interpretations compared.
FeatureCopenhagen interpretationMany-worlds interpretation
Proposed byNiels Bohr and Werner HeisenbergHugh Everett III (1957)
Before measurementelectron in a superposition of stateselectron in a superposition of states
At measurementwave function collapses to one outcomeno collapse: the universe splits, one branch for each outcome
Other outcomesvanish (not explained)happen in other worlds
Interference patternprobability of where an electron landscombined effect of all the branches
Main criticismcollapse is arbitrary; role of measurement unclearother worlds cannot be tested

Neither interpretation is universally accepted. Both try to explain the puzzling behaviour of quantum systems; the experiment itself clearly shows wave-particle duality, but the underlying reality is still being explored.

Formula sheet

Superposition
y=y_1+y_2
y_1,y_2: displacements due to each wave; y: resultant displacement
Unit: \mathrm{m}
Path difference in Young's experiment
\Delta x=d\sin\theta
d: slit separation; \theta: angle of the point from the central line
Unit: \mathrm{m}
Bright fringes (constructive)
d\sin\theta=m\lambda
m=0,\pm1,\pm2,\dots: order of the fringe
Unit: both sides in \mathrm{m}
Dark fringes (destructive)
d\sin\theta=\left(m+\tfrac12\right)\lambda
m=0,\pm1,\pm2,\dots (first minimum at m=0)
Unit: both sides in \mathrm{m}
Position of the mth bright fringe
y_m=m\,\dfrac{\lambda L}{d}
y: distance from the central maximum; L: slit-to-screen distance
Unit: \mathrm{m}
Position of a dark fringe
y=\left(m+\tfrac12\right)\dfrac{\lambda L}{d}
m=0 gives the first dark fringe
Unit: \mathrm{m}
Fringe spacing
\Delta y=\dfrac{\lambda L}{d}
\Delta y: distance between neighbouring bright (or dark) fringes
Unit: \mathrm{m}
Grating element
d=\dfrac{\text{unit length}}{\text{number of lines}}=\dfrac{1}{N}
N: number of lines per unit length (per metre)
Unit: \mathrm{m}
Grating equation
d\sin\theta=m\lambda
d: grating element; \theta: angle of the mth order maximum from the normal
Unit: both sides in \mathrm{m}
Single-slit dark bands
a\sin\theta=m\lambda
a: slit width; m=\pm1,\pm2,\dots
Unit: both sides in \mathrm{m}

Key terms

Principle of superposition
When waves overlap, the resultant displacement at any point is the sum of the individual displacements at that point.
Interference
The combination of two or more overlapping waves to give a resultant wave of larger or smaller amplitude.
Constructive interference
Interference in which the waves arrive in phase (crest on crest) and give a larger amplitude: brighter light or louder sound. Path difference =m\lambda.
Destructive interference
Interference in which the waves arrive in anti-phase (crest on trough) and give a smaller or zero amplitude: darkness or quieter sound. Path difference =\left(m+\tfrac12\right)\lambda.
Coherent sources
Sources that emit waves of the same frequency with a constant phase difference.
Monochromatic light
Light of a single wavelength (a single colour).
Path difference
The difference between the distances travelled by two waves from their sources to the same point.
Phase difference
The difference in phase between two waves at a point. It is produced by the path difference between them.
Wavelength
The distance between two consecutive crests (or troughs) of a wave, \lambda.
Frequency
The number of wave cycles passing a point per second, measured in hertz (\mathrm{Hz}).
Wavefront
A surface joining points of a wave that are in the same phase.
Fringes
The alternate bright and dark bands produced by interference or diffraction.
Fringe spacing
The distance between two consecutive bright or dark fringes, \Delta y=\lambda L/d.
Order of a fringe
The whole number m that labels a bright fringe or maximum; m=0 is the central maximum.
Antinodal and nodal lines
In a two-source pattern, lines of constructive interference (antinodal) and of destructive interference (nodal).
Beats
The regular rise and fall in loudness heard when two sounds of slightly different frequencies overlap; f_b=\lvert f_1-f_2\rvert.
Diffraction
The bending of a wave around the edges of an obstacle or through a narrow slit into the region of geometrical shadow.
Slit width
The width of the opening through which a wave passes. The smaller it is compared with the wavelength, the stronger the diffraction.
Diffraction pattern
The pattern of intensity observed after a wave diffracts around an obstacle or through a slit: alternate bright and dark bands.
Diffraction grating
A plate with a large number of closely and equally spaced parallel slits or grooves that diffracts light into sharp maxima.
Grating element (grating constant)
The distance between the centres of two adjacent slits of a grating, d=1/N.
Wave-particle duality
The fact that matter and light show both wave properties (interference, diffraction) and particle properties (arriving in single lumps).
Superposition of states
In quantum mechanics, the condition of a system existing in all its possible states at once until it is measured.
Copenhagen interpretation
The interpretation of quantum mechanics (Bohr and Heisenberg) in which a measurement collapses the wave function to one definite outcome; it stresses indeterminism, complementarity and the role of measurement.
Many-worlds interpretation
The interpretation of quantum mechanics (Everett, 1957) in which the wave function never collapses and every possible outcome happens in its own branch of the universe.

Multiple choice questionsp. 83–84

1
Rainbows are formed due to:
  1. AReflection of sunlight by water droplets
  2. BRefraction of sunlight by water dropletsAnswer
  3. CInterference of sunlight by water droplets
  4. DDiffraction of sunlight by water droplets
Why

Sunlight is refracted as it enters a raindrop, and the refractive index of water is slightly different for each colour, so the colours bend by different amounts (dispersion). The light reflects at the back of the drop and is refracted again as it leaves, which separates the colours further.

Reflection alone (A) cannot split white light into colours. Interference (C) and diffraction (D) only add faint extra bands near the violet edge of the bow; they do not make the main rainbow.

Note. The book prints the fourth option with the letter C again; it is option D here. The book's introduction says rainbows are caused by diffraction, which is not correct.

2
In the double-slit experiment, a pattern of bright and dark bands is observed on the screen. This pattern is evidence of:
  1. AReflection of light
  2. BRefraction of light
  3. CWave nature of lightAnswer
  4. DAll of the above
Why

Bright and dark bands can only be explained by the superposition of waves: crest on crest gives a bright band and crest on trough gives a dark band. Particles would give just two bright strips. Reflection and refraction play no part in forming the pattern, so A, B and D are wrong.

3
Monochromatic light from a laser passes through two slits separated by 0.00500\,\mathrm{mm}. The third bright line on a screen is formed at an angle of 18.0^\circ relative to the incident beam. What is the wavelength of the light?
  1. A51.5\,\mathrm{nm}
  2. B77.3\,\mathrm{nm}
  3. C515\,\mathrm{nm}Answer
  4. D773\,\mathrm{nm}
Why
\lambda=\dfrac{d\sin\theta}{m}=\dfrac{(5.00\times10^{-6})\sin18.0^\circ}{3}=\dfrac{(5.00\times10^{-6})(0.309)}{3}=5.15\times10^{-7}\,\mathrm{m}=515\,\mathrm{nm}

Option D (773\,\mathrm{nm}) comes from dividing by 2 instead of 3.

4
What is the width of a single slit through which 610\,\mathrm{nm} orange light passes to form a first diffraction minimum at an angle of 30^\circ?
  1. A0.863\,\mu\mathrm{m}
  2. B0.704\,\mu\mathrm{m}
  3. C0.610\,\mu\mathrm{m}
  4. D1.22\,\mu\mathrm{m}Answer
Why

For a single slit the first minimum is at a\sin\theta=\lambda.

a=\dfrac{\lambda}{\sin\theta}=\dfrac{610\times10^{-9}}{0.5}=1.22\times10^{-6}\,\mathrm{m}=1.22\,\mu\mathrm{m}

Option C (0.610\,\mu\mathrm{m}) would put the first minimum at 90^\circ, not 30^\circ.

5
Two slits are separated by a distance of 3500\,\mathrm{nm}. If light with a wavelength of 500\,\mathrm{nm} passes through the slits and produces an interference pattern, the m= ______ order minimum appears at an angle of 30^\circ.
  1. A0
  2. B1
  3. C2
  4. D3Answer
Why

Dark fringes satisfy d\sin\theta=\left(m+\tfrac12\right)\lambda:

m+\tfrac12=\dfrac{d\sin\theta}{\lambda}=\dfrac{3500\times0.5}{500}=3.5\;\Rightarrow\;m=3

In the book's numbering the first minimum has m=0, so m=3 is the fourth dark fringe from the centre.

6
What is a diffraction grating?
  1. AA single slit that produces a diffraction pattern.
  2. BA double slit that produces an interference pattern.
  3. CA periodic arrangement of slits or lines that produces a diffraction pattern.Answer
  4. DA random arrangement of slits or lines that produces a diffraction pattern.
Why

A diffraction grating has a large number of equally spaced (periodic) parallel slits or lines. A single slit (A) or a double slit (B) is not a grating, and a random arrangement (D) cannot give sharp maxima at definite angles.

7
When light travels through a narrow slit, it bends slightly around the edges. This phenomenon is called:
  1. AReflection
  2. BRefraction
  3. CDiffractionAnswer
  4. DDispersion
Why

Bending of waves around edges or through narrow openings is diffraction. Refraction is bending at the boundary between two media, and dispersion is the splitting of white light into colours.

8
A diffraction grating separates white light into its constituent colors because:
  1. AIt absorbs certain colors.
  2. BIt reflects different colors at different angles.
  3. CIt diffracts different colors at different angles.Answer
  4. DIt refracts different colors at different angles.
Why

The grating equation gives \sin\theta=m\lambda/d, so each wavelength is diffracted to its own angle (red most, violet least). A prism separates colours by refraction (D), but a grating works by diffraction and interference. It does not absorb colours (A).

92026 Annual I, Set B, Q.2 (xiv)
Light passing through double slits creates a diffraction pattern. How would the spacing of the bands in the pattern change if the slits were closer together?
  1. AThe bands would be closer together.
  2. BThe bands would spread farther apart.Answer
  3. CThe bands would remain stationary.
  4. DThe bands would fade and eventually disappear.
Why

The fringe spacing is \Delta y=\lambda L/d, so \Delta y\propto1/d. Bringing the slits closer (smaller d) makes the bands spread farther apart.

10
What is diffraction of light?
  1. AThe bending of light around an obstacle.Answer
  2. BThe reflection of light from a surface.
  3. CThe refraction of light through a medium.
  4. DThe interference of light waves.
Why

Diffraction is the bending (spreading) of light around the edges of an obstacle or through a narrow opening into the geometrical shadow. The other options describe reflection, refraction and interference.

11
The tip of a needle does not give a sharp image. It is due to
  1. APolarization
  2. BInterference
  3. CDiffractionAnswer
  4. DRefraction
Why

Light bends slightly around the very small, sharp tip of the needle, so its image or shadow has blurred edges with faint fringes instead of a sharp outline. This is diffraction.

12
What is the pattern formed on a screen when light passes through two parallel slits?
  1. AA bright central maximum with alternating dark and bright fringes.Answer
  2. BA dark central minimum with alternating bright and dark fringes.
  3. CA uniform intensity pattern.
  4. DA random pattern.
Why

At the centre the path difference from the two slits is zero, so the waves arrive in phase and the central fringe is bright. On each side come alternate dark and bright fringes.

13
What is the wavelength of light falling on double slits separated by 2.0\,\mu\mathrm{m} if the third-order maximum is at an angle of 60^\circ?
  1. A667\,\mathrm{nm}
  2. B471\,\mathrm{nm}
  3. C333\,\mathrm{nm}
  4. D577\,\mathrm{nm}Answer
Why
\lambda=\dfrac{d\sin\theta}{m}=\dfrac{(2.0\times10^{-6})(0.866)}{3}=5.77\times10^{-7}\,\mathrm{m}=577\,\mathrm{nm}

Option C (333\,\mathrm{nm}) comes from using \cos60^\circ instead of \sin60^\circ.

14
What is the longest wavelength of light passing through a single slit of width 1.20\,\mu\mathrm{m} for which there is a first-order minimum?
  1. A1.04\,\mu\mathrm{m}Answer
  2. B0.849\,\mu\mathrm{m}
  3. C0.600\,\mu\mathrm{m}
  4. D2.40\,\mu\mathrm{m}
Why

The first minimum of a single slit is at a\sin\theta=\lambda. Since \sin\theta\le1, a first minimum exists only if \lambda\le a=1.20\,\mu\mathrm{m}. Option D (2.40\,\mu\mathrm{m}) is longer than the slit width, so it gives no minimum at all.

Options A, B and C all give a first minimum (at 60^\circ, 45^\circ and 30^\circ). The longest of them is A: \lambda=a\sin60^\circ=1.20\times0.866=1.04\,\mu\mathrm{m}.

Note. Strictly, the longest wavelength is \lambda=a=1.20\,\mu\mathrm{m} (first minimum at 90^\circ), which is not among the options. The question and options are copied from OpenStax Physics; A is the best of the printed choices.

15
What is the distance between lines on a diffraction grating that produces a second-order maximum for 760\,\mathrm{nm} red light at an angle of 60^\circ?
  1. A2.28\times10^{4}\,\mathrm{nm}
  2. B3.29\times10^{2}\,\mathrm{nm}
  3. C2.53\times10^{1}\,\mathrm{nm}
  4. D1.76\times10^{3}\,\mathrm{nm}Answer
Why
d=\dfrac{m\lambda}{\sin\theta}=\dfrac{2\,(760\,\mathrm{nm})}{0.866}=1.76\times10^{3}\,\mathrm{nm}

Option B (3.29\times10^{2}\,\mathrm{nm}) comes from using \lambda\sin\theta/m, the formula turned upside down.

Short questionsp. 84–85

13 marks
How is an interference pattern formed by a diffraction grating different from the pattern formed by a double slit?
Solution

In both cases the pattern is formed by interference of light from equally spaced slits, and the bright maxima obey the same condition d\sin\theta=m\lambda. The differences come from the number of slits and their spacing:

FeatureDouble slitDiffraction grating
Number of slitstwothousands
Bright maximabroad, brightness changes graduallyvery sharp, narrow and much brighter
Between maximadark fringes as wide as the bright onesalmost completely dark
Spacingmany closely spaced fringes (small angles)a few orders, widely separated (large angles)
White lighta few coloured fringesseparate, clear spectra
Double slit and grating with the same slit spacing
Same slit spacing: the maxima are at the same angles, but the grating maxima are far sharper.

So a grating gives a few very sharp, bright lines on a dark background, while a double slit gives many broad fringes of equal width.

23 marks
A beam of light always spreads out. Why can a beam not be produced with parallel rays to prevent spreading?
Solution

Light is a wave, and every beam has to come out of an opening of finite width a, such as the window of a lamp, a lens or the output mirror of a laser. At the edges of this opening the light diffracts, so the beam spreads. The angle of spreading is roughly

\theta\approx\dfrac{\lambda}{a}\quad(\text{in radians})

The spreading would be zero only if the opening were infinitely wide, which is impossible. It can only be made small by making the beam wide compared with the wavelength. For example, a 1\,\mathrm{mm} wide beam of 633\,\mathrm{nm} light still spreads by about 6\times10^{-4}\,\mathrm{rad}.

33 marks
In the sunlight, the shadow of a building has fuzzy edges even if the building does not. Is this a refraction effect? Explain.
Solution

No. Refraction is the bending of light when it passes from one medium into another, for example from air into glass. At the edge of a building the light stays in the same medium (air), so there is no refraction.

The fuzzy edge is due to two effects. (1) Diffraction: light waves bend slightly around the edge of the building into the region of geometrical shadow, so the change from light to dark is gradual instead of sharp. (2) The Sun is not a point source but a disc about 0.5^\circ wide, so points near the edge receive light from only part of the Sun, which forms a partial shadow (penumbra).

Note. For a large building the Sun's size is actually the bigger effect, because light has such a short wavelength that diffraction blurs the edge only slightly. In this unit the expected answer is diffraction; a complete answer mentions both.

43 marks
A laser pointer emits a coherent beam of parallel light rays. Does the light from such a source spread out at all? Explain.
Solution

Yes, it spreads, but only very slightly. The beam leaves the laser through an opening about 1\,\mathrm{mm} wide, and any wave passing through an opening diffracts. The angle of spreading is about

\theta\approx\dfrac{\lambda}{a}=\dfrac{633\times10^{-9}\,\mathrm{m}}{1\times10^{-3}\,\mathrm{m}}\approx6\times10^{-4}\,\mathrm{rad}

Because the beam is very wide compared with the wavelength (a\gg\lambda), this angle is tiny, so across a room the beam looks parallel. Over long distances the spreading becomes clear: after 100\,\mathrm{m} the spot is roughly 10\,\mathrm{cm} wide. Coherence keeps the beam narrow and bright, but it cannot remove diffraction.

53 marks
A beam of light passes through a single slit to create a diffraction pattern. How will the spacing of the bands in the pattern change if the width of the slit is increased?
Solution

The spacing of the bands decreases: the bands move closer together and the central bright band becomes narrower (and brighter, because more light passes through).

The dark bands of a single slit are at a\sin\theta=m\lambda, so \sin\theta=m\lambda/a. A wider slit (larger a) gives smaller angles for every band. For small angles the width of the central band on a screen at distance L is

w=\dfrac{2\lambda L}{a}\;\propto\;\dfrac{1}{a}

So doubling the slit width halves the spacing of the bands. If the slit becomes much wider than the wavelength, the bands crowd together and we simply see a sharp image of the slit.

63 marks
Describe a diffraction grating and the interference pattern it produces.
Solution

A diffraction grating is a plate of glass or plastic with a very large number of equally spaced, parallel slits (lines), for example 5000 lines per centimetre. The distance between neighbouring slits is the grating element d=1/N, where N is the number of lines per unit length.

When monochromatic light falls normally on it, the waves from all the slits interfere. They reinforce only in directions where the path difference between neighbouring slits is a whole number of wavelengths:

d\sin\theta=m\lambda,\qquad m=0,1,2,\dots

The pattern is a bright central maximum (m=0) with a few very sharp, bright lines (first order, second order, ...) on both sides, separated by almost complete darkness. The number of orders is limited by m\le d/\lambda. With white light the central maximum is white and each order is a spectrum from violet (inside) to red (outside).

Sharp orders produced by a grating with monochromatic light.
72026 Annual I, Set B, MCQ 62026 Annual I, Set D, Q.2 (xiii)3 marks
Suppose a monochromatic light falls on a diffraction grating. What happens to the interference pattern if the same light falls on a grating that has more lines per centimeter?
Solution

More lines per centimetre means a smaller grating element d=1/N. From the grating equation

\sin\theta=\dfrac{m\lambda}{d}

a smaller d gives a larger angle for every order. So (1) the maxima spread farther apart; (2) fewer orders can be seen, because \sin heta=m\lambda/d cannot be more than 1; (3) the maxima become sharper, because more slits share the beam.

For example, with 600\,\mathrm{nm} light a grating of 2000 lines per cm (d=5\,\mu\mathrm{m}) gives the first order at 6.9^\circ and up to 8 orders, while a grating of 6000 lines per cm (d=1.67\,\mu\mathrm{m}) gives the first order at 21.1^\circ and only 2 orders.

83 marks
What is the significance of the equation d\sin\theta=m\lambda in the context of (a) diffraction gratings (b) Young's double-slit experiment?
Solution
(a) Diffraction gratings

Here d is the grating element (distance between neighbouring slits) and \theta is the angle of a sharp principal maximum. The equation gives the angles at which the bright lines (orders) appear. It is used to measure the wavelength, \lambda=d\sin\theta/m, by measuring \theta for a known order. It also gives the number of orders possible (m\le d/\lambda), and it shows that different wavelengths are diffracted to different angles, which is how a grating separates white light into a spectrum.

(b) Young's double-slit experiment

Here d is the separation of the two slits. The equation is the condition for a bright fringe: the path difference d\sin\theta equals a whole number of wavelengths. With the small-angle approximation \sin\theta\approx y/L it gives the fringe positions y=m\lambda L/d and the fringe spacing \Delta y=\lambda L/d, which is used to find \lambda. Dark fringes satisfy d\sin\theta=\left(m+\tfrac12\right)\lambda instead.

93 marks
What is the effect of following aspects of diffraction grating on the resulting interference pattern: (a) number of slits (b) width of the slits?
Solution
(a) Number of slits

Increasing the number of slits (with the same spacing d) does not move the maxima, because their angles are fixed by d\sin\theta=m\lambda. It makes the maxima narrower (sharper) and brighter, and the regions between them darker, so the grating can separate wavelengths that are very close together. (If the number of slits per unit length is increased, d becomes smaller and the maxima also spread farther apart.)

(b) Width of the slits

The width a of each slit decides how widely each slit spreads its light (by an angle of about \lambda/a). Narrow slits spread light widely, so many orders are bright. Wider slits keep most of the light near the centre, so the higher orders become faint. The slit width does not change the positions of the maxima, only their brightness; wider slits also let more light through, so the pattern as a whole is brighter.

102026 Annual I, Set D, Q.2 (ii) OR3 marks
Describe the conditions necessary for sustained interference patterns to be observed in Young's double-slit experiment.
Solution

A sustained pattern is one that stays fixed in position, so that it can be seen and measured. In Young's experiment this needs:

  1. Coherent sources: the two slits must give waves of the same frequency with a constant phase difference. In Young's experiment both slits are lit by the same wavefront from one single slit; two separate lamps cannot be used.
  2. Monochromatic light: light of one wavelength. Otherwise fringes of different colours (different spacings) overlap and blur.
  3. Narrow slits, close together, and a distant screen: narrow slits diffract the light so the two beams overlap, and a small d with a large L makes the fringe spacing \Delta y=\lambda L/d large enough to see.
  4. Equal amplitudes: slits of equal width, equally lit, so that the dark fringes are completely dark.
  5. A stable set-up: no vibrations or air currents that would shift the fringes.

Comprehensive questionsp. 85

1
Design an experiment using a ripple tank to demonstrate the principles of interference with water waves. Describe what you expect to observe when two wave sources are introduced.
Solution
Aim

To show interference of water waves produced by two coherent point sources.

Apparatus
  • A ripple tank: a shallow tray with a glass bottom, holding water about 1\,\mathrm{cm} deep, with sloping sponge edges to stop reflections.
  • A vibrator (small electric motor) on a bar carrying two point dippers.
  • A lamp above the tank and a sheet of white paper (screen) below it.
  • A hand stroboscope (optional) to freeze the pattern.
Ripple tank arrangement with two dippers driven by one motor.
Procedure
  1. Level the tank and fill it with water to a depth of about 1\,\mathrm{cm}.
  2. Fix the two dippers on the vibrating bar a few centimetres apart, just touching the water.
  3. Switch on the lamp and the vibrator. Both dippers are driven by the same motor, so they vibrate with the same frequency and in phase: they are coherent sources.
  4. Watch the shadow pattern on the white paper, freeze it with the stroboscope if needed, and sketch it.
  5. Change the frequency of the vibrator and the separation of the dippers, and note how the pattern changes.
Observations
Expected pattern (top view): antinodal lines (red) and nodal lines (dashed) spread out from between the two sources.
  • Circular waves spread out from each dipper and overlap.
  • Along certain lines, antinodal lines, crest meets crest and trough meets trough: the water moves up and down strongly and the paper shows sharp bright and dark bands.
  • Between them lie nodal lines, where crest meets trough: the water is almost still and the paper shows an even grey.
  • The lines fan out from the region between the sources like curved spokes (hyperbolas). The central line, equidistant from both sources, is antinodal.
Explanation

At a point P the waves from S_1 and S_2 have travelled different distances. The type of interference depends on this path difference:

S_2P-S_1P=m\lambda\ \ (\text{antinodal}),\qquad S_2P-S_1P=\left(m+\tfrac12\right)\lambda\ \ (\text{nodal})
Effect of changing the conditions
  • Higher frequency (shorter wavelength): more lines, closer together.
  • Greater separation of the dippers: more lines, closer together.
  • Only one dipper working: plain circular waves with no nodal lines, so no interference.
2
Briefly explain why we see a spectrum of colors (red, orange, yellow, green, blue, indigo, violet) in a rainbow. Is it due to reflection, refraction, or interference of light?
Solution

A rainbow is produced mainly by refraction with dispersion of sunlight in raindrops, together with one reflection inside each drop. It is not caused by interference.

What happens inside a raindrop
  1. Refraction and dispersion on entering. Sunlight is white light, a mixture of colours. When it enters a spherical raindrop it slows down and bends. The refractive index of water is slightly larger for violet light (n\approx1.343) than for red light (n\approx1.331), so violet bends more than red and the colours begin to separate.
  2. Reflection at the back of the drop. The light is reflected from the inner back surface of the drop.
  3. Refraction on leaving. The light refracts again as it comes out, which separates the colours further. Red light leaves at about 42^\circ to the direction of the incoming sunlight and violet at about 40^\circ.
Path of sunlight in a raindrop: refraction with dispersion, reflection at the back, and refraction again.
Why we see an arc of colours

Each drop sends only one colour into the observer's eye. Drops that lie about 42^\circ from the point directly opposite the Sun send red light, and drops at about 40^\circ send violet light. All the drops at the same angle form a circle around that point, so we see circular bands of colour, red on the outside and violet on the inside, with the other colours in between. The ground cuts off the lower part, so we usually see an arc. The Sun must be behind the observer.

Reflection, refraction or interference
ProcessRole in the rainbow
Refraction (with dispersion)main cause: separates the colours
Reflectionsends the light back towards the observer, but alone cannot separate colours
Interferenceonly faint extra bands just inside the violet edge; not the main colours

A fainter secondary rainbow, at about 51^\circ, is formed by two reflections inside the drops; its colours are in the reverse order.

Note. The book's introduction says rainbows come from diffraction in water droplets. The correct explanation is refraction with dispersion plus reflection inside the drops.

3
Describe two examples of how interference of light waves plays a role in natural phenomena you can observe in everyday life (excluding rainbows).
Solution

Two common examples are the colours of thin films (soap bubbles and oil on a wet road) and the shining colours of peacock feathers, butterfly wings and pearl shells. In both, one light wave is split into two by reflection at two surfaces, so the two reflected waves are coherent and can interfere.

1. Colours of soap bubbles and oil films
Interference in a thin film: ray 1 reflects from the top surface and ray 2 from the bottom surface.

When white light falls on a thin film of soap or oil, part of it reflects from the top surface (ray 1) and part enters the film, reflects from the bottom surface and comes out again (ray 2). Ray 2 travels an extra distance of about twice the film thickness t. For a given thickness, this path difference makes one colour interfere constructively (bright) while another colour interferes destructively (missing), so that part of the film looks coloured.

The thickness changes from place to place, so we see bands of different colours. As a soap bubble drains and becomes thinner, the colours move and change; just before it bursts the film is so thin that it looks black. The exact condition for a bright reflection includes a half-wavelength shift at the top surface: 2nt=\left(m+\tfrac12\right)\lambda, where n is the refractive index of the film.

2. Peacock feathers, butterfly wings and pearl shells

The brilliant blue of a peacock feather or of a Morpho butterfly's wing does not come from a pigment. These surfaces contain many very thin, regularly spaced transparent layers. Light reflected from the different layers interferes: one colour (often blue or green) is strongly reinforced and the others cancel. The colour changes as you look from a different angle, because the path difference depends on the angle. The shiny inside of pearl and oyster shells (mother-of-pearl) works in the same way.

4
Sometimes, when two tuning forks with slightly different frequencies are struck close together, you hear a wavering sound instead of a clear tone. Explain what causes this 'beat' phenomenon, and how is it related to the interference of sound waves?
Solution
Cause of beats

The two tuning forks send out sound waves of slightly different frequencies f_1 and f_2. Both waves reach the ear at the same time and, by the principle of superposition, the ear receives their sum. Because the frequencies are different, the phase difference between the two waves changes steadily with time.

  • At one instant the two waves are in phase (compression with compression): they interfere constructively and the sound is loud.
  • A little later one fork has gained half a vibration on the other, so the waves are in anti-phase (compression with rarefaction): they interfere destructively and the sound is soft.
  • Later still they are back in phase and the sound is loud again.

This regular rise and fall of loudness is the wavering sound called beats.

Resultant of waves of 10\,\mathrm{Hz} and 12\,\mathrm{Hz}: the loudness rises and falls twice per second (2 beats per second).

The number of beats heard per second equals the difference between the two frequencies: forks of 256\,\mathrm{Hz} and 260\,\mathrm{Hz} give 4 beats per second.

Relation to interference

Beats are interference of sound waves in time, heard at one place, while the two-loudspeaker pattern is interference in space (loud and quiet places). Both follow from the superposition of two waves. For a steady interference pattern in space the sources must be coherent (equal frequencies); when the frequencies differ slightly, the interference is not steady but changes with time, and we hear beats.

Uses
  • Tuning musical instruments: a string is adjusted until the beats with a standard fork disappear.
  • Finding an unknown frequency: count the beats with a fork of known frequency.
5
Both CDs and DVDs use lasers to store information, but DVDs hold much more data. How might the principles of diffraction be used in DVD design to achieve this higher storage capacity?
Solution

CDs and DVDs store data as a spiral track of microscopic pits in a reflecting layer. A laser is focused to a tiny spot on the track, and a detector reads the reflected light. The amount of data depends on how small the pits can be and how close together the turns of the track can be placed. This is limited by diffraction.

Diffraction limits the size of the laser spot

A lens cannot focus light to a true point. Because light diffracts at the edge of the lens, the smallest spot has a diameter roughly proportional to \lambda/\mathrm{NA}, where \lambda is the wavelength and NA (numerical aperture) measures how wide the cone of light from the lens is. A smaller spot can read smaller pits that are packed closer together without reading the neighbouring tracks as well.

How DVD design uses this
  • Shorter wavelength: a DVD uses a red laser of 650\,\mathrm{nm} instead of the CD's infrared laser of 780\,\mathrm{nm}.
  • Wider lens cone: the DVD lens has NA =0.60 instead of 0.45.
  • Together these make the spot area only about 0.4 of a CD's, since \left(\dfrac{650/0.60}{780/0.45}\right)^2\approx0.39. So the tracks can be brought closer (0.74\,\mu\mathrm{m} instead of 1.6\,\mu\mathrm{m}) and the pits made shorter (0.40\,\mu\mathrm{m} instead of 0.83\,\mu\mathrm{m}).
  • A DVD can also have two data layers; the laser is refocused to read the second layer.
CDDVDBlu-ray
Laser wavelength780\,\mathrm{nm} (infrared)650\,\mathrm{nm} (red)405\,\mathrm{nm} (blue-violet)
Numerical aperture0.450.600.85
Track spacing1.6\,\mu\mathrm{m}0.74\,\mu\mathrm{m}0.32\,\mu\mathrm{m}
Capacity (one layer)about 0.7 GB4.7 GB25 GB
Interference in reading the data

The pits are about a quarter of a wavelength deep (measured in the plastic). Light reflected from the bottom of a pit travels half a wavelength farther than light reflected from the flat surface around it, so the two interfere destructively and the reflected signal drops. The detector reads these changes as the data.

6
Describe a diffraction grating and its key features like grating constant and slit width. How does a diffraction grating separate white light into its constituent colors?
Solution
Diffraction grating

A diffraction grating is a plate of glass or plastic on which a very large number of equally spaced, parallel lines are ruled, for example 600 lines per millimetre. The ruled lines are opaque, and the clear gaps between them act as slits. When light falls on the grating, every slit diffracts it, and the waves from all the slits interfere to give sharp bright maxima in certain directions.

Key features
  • Grating constant (grating element) d: the distance between the centres of two neighbouring slits, d=a+b (slit width plus width of an opaque line). If there are N lines per metre, d=1/N. For 600 lines per mm, d=1.67\times10^{-6}\,\mathrm{m}. It fixes the angles of the maxima and must be larger than \lambda.
  • Slit width a: the width of each clear gap, smaller than d. Each slit spreads its light by an angle of about \lambda/a; narrow slits spread the light widely so that many orders are bright.
  • Number of lines: the more lines the beam covers, the sharper and brighter the maxima, and the better the grating separates close wavelengths.
Grating equation

Derivation d\sin\theta=m\lambda

Aim. Find the directions of the bright maxima.

Path difference d\sin\theta between neighbouring slits.
  1. Light falls normally on the grating, so all slits are in phase. Consider rays leaving neighbouring slits at angle \theta; the ray from the lower slit travels farther by the side opposite \theta in a right triangle with hypotenuse d.
    \Delta x=d\sin\theta
  2. Every slit is d\sin\theta behind its neighbour, so if this is a whole number of wavelengths, the waves from all the slits arrive in phase and reinforce.
    d\sin\theta=m\lambda,\quad m=0,1,2,\dots
  3. In all other directions the many waves cancel, so the space between the maxima is dark.
Resultd\sin\theta=m\lambda

For a given order, the angle depends on the wavelength: \sin\theta=m\lambda/d.

Separation of white light

White light contains all wavelengths from violet (about 400\,\mathrm{nm}) to red (about 700\,\mathrm{nm}). At \theta=0 (m=0) every wavelength is in phase, so the central maximum is white. In each other order, each wavelength is diffracted to its own angle given by \sin\theta=m\lambda/d. Violet has the shortest wavelength and is deviated least; red has the longest and is deviated most. So each order spreads into a spectrum, violet on the inside and red on the outside.

White light through a grating with 600 lines per mm: violet at 13.9^\circ and red at 24.8^\circ in the first order.

For 600 lines per mm, the first-order spectrum runs from \sin^{-1}(0.40/1.667)=13.9^\circ for violet to \sin^{-1}(0.70/1.667)=24.8^\circ for red. Higher orders are more spread out, and the second and third orders overlap. Unlike a prism, which deviates violet most, a grating deviates red most.

7
Scientists use X-ray diffraction to determine the structure of crystals. Briefly explain how the diffraction pattern of X-rays helps reveal the arrangement of atoms within a crystal lattice.
Solution
Why X-rays and crystals

Diffraction is strong only when the spacing of the diffracting objects is similar to the wavelength. X-rays have wavelengths of about 10^{-10}\,\mathrm{m}, the same order as the distance between atoms in a crystal (1 to 3\times10^{-10}\,\mathrm{m}). A crystal is a regular three-dimensional arrangement of atoms, so it acts as a three-dimensional diffraction grating for X-rays. Visible light has far too long a wavelength to do this.

How the pattern reveals the arrangement of atoms
  1. A narrow beam of X-rays of known wavelength is sent through a small crystal, and the diffracted beams are recorded on a film or electronic detector as a pattern of spots.
  2. The positions (angles) of the spots give the spacings d between the different sets of atomic planes, and so the size and shape of the repeating unit of the lattice.
  3. The symmetry of the spot pattern shows the symmetry of the crystal (for example cubic or hexagonal).
  4. The brightness of each spot depends on which atoms lie in the planes and how many electrons they have, so the intensities show where each kind of atom sits.
  5. Combining all the spots mathematically (by computer) gives a three-dimensional map of the atoms in the crystal.
Examples

This method revealed the arrangement of sodium and chloride ions in common salt. X-ray diffraction photographs of DNA taken by Rosalind Franklin helped Watson and Crick discover its double-helix structure in 1953, and today the method is used to find the structures of proteins and new medicines.

Numerical problemsp. 85

Convert every length to metres first: 1\,\mathrm{nm}=10^{-9}\,\mathrm{m}, 1\,\mu\mathrm{m}=10^{-6}\,\mathrm{m}, 1\,\mathrm{mm}=10^{-3}\,\mathrm{m}.

1
The light with wavelength (\lambda) 633\,\mathrm{nm} is used at a distance of 2\,\mathrm{m} from the screen. The separation between the slits is 10\,\mu\mathrm{m}. Find the distance between adjacent maxima (fringe width) for the first (m=1) and second (m=2) order of interference.
Solution
Given
  • \lambda = 633\,\mathrm{nm}=6.33\times10^{-7}\,\mathrm{m}
  • L = 2\,\mathrm{m}
  • d = 10\,\mu\mathrm{m}=1.0\times10^{-5}\,\mathrm{m}
To find
  • fringe width \Delta y for m=1 and m=2
  1. The distance between adjacent maxima is the fringe spacing. It does not contain m, so it is the same for every order.
    \Delta y=\dfrac{\lambda L}{d}
  2. Substitute the values in SI units.
    \Delta y=\dfrac{(6.33\times10^{-7})(2)}{1.0\times10^{-5}}=0.1266\,\mathrm{m}
  3. Check with the fringe positions y_m=m\lambda L/d: the first- and second-order maxima are at 0.127\,\mathrm{m} and 0.253\,\mathrm{m}, and the gap between them is again 0.127\,\mathrm{m}.
    y_2-y_1=0.2532-0.1266=0.1266\,\mathrm{m}
  4. Check the small-angle approximation: \sin\theta_2=2\lambda/d=0.127, so \theta_2=7.3^\circ, which is small. The exact positions differ from these by less than 1\%.
\Delta y=0.127\,\mathrm{m}\ (127\,\mathrm{mm})\ \text{for both } m=1 \text{ and } m=2

Book answer. 0.127\,\mathrm{mm} (first order) and 0.063\,\mathrm{mm} (second order)

Note. The book's answers are wrong. With d=10\,\mu\mathrm{m} the fringe width is 0.127\,\mathrm{m}, not 0.127\,\mathrm{mm} (the book's value would need d=10\,\mathrm{mm}). The fringe width is the same for all orders; the book's 0.063\,\mathrm{mm} is half of its own fringe width, which is the distance from the central maximum to the first dark fringe.

2
Find the distance from the central maximum to the third minimum on the screen, if the light of wavelength 400\,\mathrm{nm} is used. The distance between the slits and screen is 0.5\,\mathrm{m} and slit separation is 5\,\mu\mathrm{m}.
Solution
Given
  • \lambda = 400\,\mathrm{nm}=4.00\times10^{-7}\,\mathrm{m}
  • L = 0.5\,\mathrm{m}
  • d = 5\,\mu\mathrm{m}=5.0\times10^{-6}\,\mathrm{m}
To find
  • distance y of the third dark fringe from the centre
  1. Dark fringes are at y=\left(m+\tfrac12\right)\lambda L/d. The first minimum has m=0, the second m=1, so the third minimum has m=2.
    y=\left(2+\tfrac12\right)\dfrac{\lambda L}{d}
  2. Work out the fringe spacing first.
    \dfrac{\lambda L}{d}=\dfrac{(4.00\times10^{-7})(0.5)}{5.0\times10^{-6}}=0.040\,\mathrm{m}
  3. The third dark fringe is two and a half fringe spacings from the centre.
    y=2.5\times0.040=0.10\,\mathrm{m}
  4. Check: \sin\theta=2.5\lambda/d=0.2, so \theta=11.5^\circ. The exact value L\tan\theta=0.102\,\mathrm{m} is almost the same.
y=0.10\,\mathrm{m}\ (10\,\mathrm{cm})

Book answer. 0.024\,\mathrm{mm}

Note. The book's answer cannot be obtained from the given data. With these values the fringe spacing alone is 0.040\,\mathrm{m}, and the third dark fringe is 2.5 fringe spacings from the centre, 0.10\,\mathrm{m}.

3
The distance between the slits and screen is 1\,\mathrm{m}, fringe spacing on screen is 1\,\mathrm{mm}. Find the slit separation when light wavelength is 500\,\mathrm{nm}.
Solution
Given
  • L = 1\,\mathrm{m}
  • \Delta y = 1\,\mathrm{mm}=1.0\times10^{-3}\,\mathrm{m}
  • \lambda = 500\,\mathrm{nm}=5.00\times10^{-7}\,\mathrm{m}
To find
  • slit separation d
  1. Start from the fringe-spacing formula and make d the subject.
    \Delta y=\dfrac{\lambda L}{d}\;\Rightarrow\;d=\dfrac{\lambda L}{\Delta y}
  2. Substitute.
    d=\dfrac{(5.00\times10^{-7})(1)}{1.0\times10^{-3}}=5.0\times10^{-4}\,\mathrm{m}
  3. Check: with d=0.5\,\mathrm{mm}, \Delta y=(5\times10^{-7})(1)/(5\times10^{-4})=1\times10^{-3}\,\mathrm{m}, as given.
d=5.0\times10^{-4}\,\mathrm{m}=0.50\,\mathrm{mm}

Book answer. 0.05\,\mathrm{mm}

Note. The book's answer is 10 times too small. With d=0.05\,\mathrm{mm} the fringe spacing would be 10\,\mathrm{mm}, not 1\,\mathrm{mm}.

4
Find the minimum angle for non-zero diffracted light, having slit spacing 2.0\,\mu\mathrm{m} with 400\,\mathrm{nm} light wavelength.
Solution
Given
  • d = 2.0\,\mu\mathrm{m}=2.0\times10^{-6}\,\mathrm{m}
  • \lambda = 400\,\mathrm{nm}=4.00\times10^{-7}\,\mathrm{m}
To find
  • smallest angle \theta of a diffracted beam with m\neq0
  1. Light at \theta=0 is the central (zero-order) maximum. The nearest bright diffracted beam with m\neq0 is the first order, m=1.
    d\sin\theta=m\lambda,\quad m=1
  2. Solve for \sin\theta.
    \sin\theta=\dfrac{\lambda}{d}=\dfrac{4.00\times10^{-7}}{2.0\times10^{-6}}=0.20
  3. Take the inverse sine.
    \theta=\sin^{-1}(0.20)=11.54^\circ
\theta_{\min}=11.54^\circ\ (\text{first-order maximum})

Book answer. \theta_{\min}=11.54^\circ (the book calls it the minimum angle for first minimum)

Note. The book's number is right, but it is the angle of the first-order maximum, not of a minimum. The first dark fringe for this slit spacing would be at \sin\theta=\lambda/2d=0.10, that is 5.74^\circ.

5
The deviation of second order diffracted image formed by an optical grating having 5000 lines per centimeter is 32^\circ. Calculate the wavelength of light used.
Solution
Given
  • N = 5000\ \text{lines per cm}=5.0\times10^{5}\,\mathrm{m^{-1}}
  • m = 2
  • \theta = 32^\circ
To find
  • wavelength \lambda
  1. Grating element from the number of lines per unit length.
    d=\dfrac{1}{N}=\dfrac{1}{5.0\times10^{5}}=2.0\times10^{-6}\,\mathrm{m}
  2. Grating equation, solved for \lambda.
    \lambda=\dfrac{d\sin\theta}{m}
  3. Substitute; \sin32^\circ=0.530.
    \lambda=\dfrac{(2.0\times10^{-6})(0.530)}{2}=5.3\times10^{-7}\,\mathrm{m}
  4. Convert to centimetres to compare with the book.
    5.3\times10^{-7}\,\mathrm{m}=5.3\times10^{-5}\,\mathrm{cm}
\lambda=5.3\times10^{-7}\,\mathrm{m}=5.3\times10^{-5}\,\mathrm{cm}\ (530\,\mathrm{nm})

Book answer. 5.3\times10^{-5}\,\mathrm{cm} (agrees)

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 61 mark
If the number of lines per unit length in a grating increase, the diffraction pattern (maxima) becomes:
  1. ALess sharp and closely spaced
  2. BSharper and widely spacedAnswer
  3. CMore widely spaced and less sharp
  4. DMore closely spaced and sharper
Why

More lines per unit length means a smaller grating element d=1/N. From \sin\theta=m\lambda/d every order moves to a larger angle, so the maxima are more widely spaced. With more lines sharing the beam the maxima are also narrower and brighter, that is sharper. Options C and D each get only one of the two changes right.

2026 Annual I, Set B, Q.2 (ii) OR1.5+1.5 = 3 marks
Briefly explain constructive and destructive interference of light. Give mathematical expression for both in terms of path difference.
Solution
Constructive interference (1.5 marks)

When two coherent light waves meet in phase, crest on crest and trough on trough, their displacements add and the resultant wave has a larger amplitude. The light there is brighter: a bright fringe. This happens when the path difference is a whole number of wavelengths:

\Delta x=S_2P-S_1P=m\lambda,\qquad m=0,\pm1,\pm2,\dots
Destructive interference (1.5 marks)

When two coherent light waves meet in anti-phase, crest on trough, they cancel and the resultant amplitude is small or zero. The point is dark: a dark fringe. This happens when the path difference is an odd number of half wavelengths:

\Delta x=\left(m+\tfrac12\right)\lambda=\dfrac{\lambda}{2},\ \dfrac{3\lambda}{2},\ \dfrac{5\lambda}{2},\ \dots

In Young's double slit the path difference to a point at angle \theta is d\sin\theta, so bright fringes need d\sin\theta=m\lambda and dark fringes need d\sin\theta=\left(m+\tfrac12\right)\lambda.

2026 Annual I, Set B, Q.2 (xiii)2+1 = 3 marks
What is meant by grating element of a diffraction grating? Also write its formula.
Solution
Grating element (2 marks)

A diffraction grating is a plate with a very large number of equally spaced, parallel slits. The grating element (grating constant) d is the distance between the centres of two adjacent slits. It equals the width of one transparent slit a plus the width of one opaque line b, so d=a+b. It is a few micrometres, a little larger than the wavelength of light, and its SI unit is the metre.

Formula (1 mark)
d=\dfrac{\text{unit length of grating}}{\text{number of lines ruled on it}}=\dfrac{1}{N}

For example, a grating with 5000 lines per cm has d=\dfrac{1\times10^{-2}\,\mathrm{m}}{5000}=2.0\times10^{-6}\,\mathrm{m}. The grating element appears in the grating equation d\sin\theta=m\lambda.

2026 Annual I, Set B, Q.2 (xiv)2+1 = 3 marks
How does increasing slit separation affect fringe spacing? Discuss with mathematical relation.
Solution
Effect (2 marks)

Increasing the slit separation d decreases the fringe spacing: the bright and dark fringes come closer together. With the slits farther apart, the path difference d\sin\theta grows faster as we move across the screen, so it reaches one more wavelength after a shorter distance. If d is made too large, the fringes are so close that they cannot be seen separately.

Mathematical relation (1 mark)
\Delta y=\dfrac{\lambda L}{d}\quad\Rightarrow\quad\Delta y\propto\dfrac{1}{d}\quad(\lambda,\ L\ \text{constant})

For example, doubling d halves the fringe spacing: with \lambda=600\,\mathrm{nm} and L=1\,\mathrm{m}, d=0.25\,\mathrm{mm} gives \Delta y=2.4\,\mathrm{mm}, while d=0.50\,\mathrm{mm} gives \Delta y=1.2\,\mathrm{mm}.

2026 Annual I, Set B, Q.44+1+2 = 7 marks
Explain the working of Young's double slit experiment. Make diagram. Also derive the expression for fringe spacing.
Solution
Working of Young's double-slit experiment (4 marks)

Young's experiment (1801) produces interference of light from two coherent sources and shows that light behaves as a wave.

  1. Monochromatic light (for example from a sodium lamp or a laser) falls on a narrow single slit S_0. Light spreading out from S_0 by diffraction reaches two narrow, parallel slits S_1 and S_2, placed close together (separation d, about 0.5\,\mathrm{mm}) and at equal distances from S_0.
  2. S_1 and S_2 receive light from the same wavefront, so they act as two coherent sources vibrating in phase.
  3. Each slit diffracts the light, and the two sets of waves overlap and interfere in the region beyond the slits. A screen is placed at a distance L (about 1\,\mathrm{m}) from the slits.
  4. At points where the path difference S_2P-S_1P is 0,\ \lambda,\ 2\lambda,\dots the waves arrive in phase and interfere constructively, giving bright fringes.
  5. At points where the path difference is \tfrac12\lambda,\ \tfrac32\lambda,\dots the waves arrive in anti-phase and interfere destructively, giving dark fringes.
  6. So equally spaced, alternate bright and dark fringes parallel to the slits appear on the screen. The central fringe at O is bright because the path difference there is zero. With white light the central fringe is white and the others are coloured.
Diagram (1 mark)
Young's double-slit experiment: source, single slit S_0, double slit S_1S_2 (separation d), screen at distance L with bright and dark fringes.
Derivation of fringe spacing (2 marks)

Derivation Fringe spacing \Delta y=\lambda L/d

Aim. Find the distance between two neighbouring bright fringes.

Assumptions.
  • The screen is far away (L\gg d) and the angle \theta is small.
Path difference BC =d\sin\theta for a point P at height y above the centre O.
  1. Path difference. Draw S_1C perpendicular to S_2P. For a distant screen the rays are nearly parallel, so the path difference is BC. In the right triangle S_1S_2C the angle at S_1 equals \theta and the hypotenuse is d.
    S_2P-S_1P=BC=d\sin\theta
  2. Bright fringe. For the mth bright fringe the path difference is m whole wavelengths.
    d\sin\theta=m\lambda
  3. Position. \theta is small, so \sin\theta\approx\tan\theta=y_m/L, where y_m is the distance of the fringe from O.
    d\,\dfrac{y_m}{L}=m\lambda\;\Rightarrow\;y_m=\dfrac{m\lambda L}{d}
  4. Spacing. The next bright fringe is at y_{m+1}=(m+1)\lambda L/d. Subtract.
    \Delta y=y_{m+1}-y_m=\dfrac{(m+1)\lambda L}{d}-\dfrac{m\lambda L}{d}=\dfrac{\lambda L}{d}
Result\Delta y=\dfrac{\lambda L}{d}

The fringes are equally spaced. The spacing increases with the wavelength and the screen distance, and decreases when the slit separation increases. Dark fringes, at y=\left(m+\tfrac12\right)\lambda L/d, have the same spacing.

\Delta y=\dfrac{\lambda L}{d}
2026 Annual I, Set D, MCQ 81 mark
The wavelength of light can be accurately determined by using a:
  1. AInterferometer
  2. BCapacitor
  3. CMelde's apparatus
  4. DDiffraction gratingAnswer
Why

A diffraction grating gives very sharp, widely separated maxima whose angles can be measured accurately. With the grating element d known, \lambda=d\sin\theta/m (book section 18.2.1). A capacitor stores charge, and Melde's apparatus studies standing waves on a string.

Note. An interferometer (such as Michelson's) can also measure the wavelength of light very precisely, so option A is not wrong in itself. The new book presents the diffraction grating as the device for this, so D is the expected answer.

2026 Annual I, Set D, Q.2 (ii) OR1x3 = 3 marks
State the conditions necessary for observing interference fringes.
Solution

Any three of the following (1 mark each):

  1. Coherent sources: the two sources must have the same frequency and a constant phase difference. (In Young's experiment both slits are lit by the same wavefront from one single slit.)
  2. Monochromatic light: light of a single wavelength; otherwise fringes of different colours overlap and the pattern blurs.
  3. Narrow, closely spaced slits and a distant screen: each slit must be narrow enough to diffract the light so that the two beams overlap, and d must be small and L large so that the fringe spacing \Delta y=\lambda L/d is big enough to see.
  4. Nearly equal amplitudes: the two waves should be of equal intensity so that the dark fringes are completely dark.
  5. Stable arrangement: no vibrations or air currents that would shift the fringes.
2026 Annual I, Set D, Q.2 (xiii)2+1 = 3 marks
How does increasing the number of slits per unit length affect the diffraction pattern? Make a sketch.
Solution
Effect (2 marks)

Increasing the number of slits per unit length N decreases the grating element d=1/N. From the grating equation \sin\theta=m\lambda/d:

  • each order appears at a larger angle, so the maxima are more widely spaced;
  • fewer orders can be seen, because \sin heta=m\lambda/d cannot be more than 1;
  • the maxima become sharper and brighter, because more slits share the same beam.
Sketch (1 mark)
Effect of more lines per unit length
With more lines per mm (d reduced from 3.6\lambda to 1.5\lambda) the first order moves from 16.1^\circ to 41.8^\circ (\sin\theta=m\lambda/d). The maxima are more widely spaced and fewer orders fit in: 3 bright lines instead of 7. Each maximum is narrow compared with the gap between maxima.
2026 Annual I, Set D, Q.2 (xiv)2+1 = 3 marks
How can one increase the fringe spacing in Young's double slit experiment? Also write its mathematical calculation.
Solution
Ways to increase the fringe spacing (2 marks)

The fringe spacing is \Delta y=\lambda L/d, so it can be increased by:

  • moving the screen farther from the slits (increasing L);
  • using light of a longer wavelength, for example red instead of blue (increasing \lambda);
  • bringing the two slits closer together (decreasing d).
Mathematical calculation (1 mark)
\Delta y=\dfrac{\lambda L}{d}=\dfrac{(6.0\times10^{-7}\,\mathrm{m})(1.0\,\mathrm{m})}{5.0\times10^{-4}\,\mathrm{m}}=1.2\times10^{-3}\,\mathrm{m}

With \lambda=600\,\mathrm{nm}, L=1.0\,\mathrm{m} and d=0.50\,\mathrm{mm} the spacing is 1.2\,\mathrm{mm}. Doubling L to 2.0\,\mathrm{m}, or halving d to 0.25\,\mathrm{mm}, doubles it to 2.4\,\mathrm{mm}.

2026 Annual I, Set D, Q.44+1+2 = 7 marks
Explain the working of Young's double slit experiment. Make diagram. Also derive the expression for fringe spacing.
Solution
Working of Young's double-slit experiment (4 marks)

Young's experiment (1801) produces interference of light from two coherent sources and shows that light behaves as a wave.

  1. Monochromatic light (for example from a sodium lamp or a laser) falls on a narrow single slit S_0. Light spreading out from S_0 by diffraction reaches two narrow, parallel slits S_1 and S_2, placed close together (separation d, about 0.5\,\mathrm{mm}) and at equal distances from S_0.
  2. S_1 and S_2 receive light from the same wavefront, so they act as two coherent sources vibrating in phase.
  3. Each slit diffracts the light, and the two sets of waves overlap and interfere in the region beyond the slits. A screen is placed at a distance L (about 1\,\mathrm{m}) from the slits.
  4. At points where the path difference S_2P-S_1P is 0,\ \lambda,\ 2\lambda,\dots the waves arrive in phase and interfere constructively, giving bright fringes.
  5. At points where the path difference is \tfrac12\lambda,\ \tfrac32\lambda,\dots the waves arrive in anti-phase and interfere destructively, giving dark fringes.
  6. So equally spaced, alternate bright and dark fringes parallel to the slits appear on the screen. The central fringe at O is bright because the path difference there is zero. With white light the central fringe is white and the others are coloured.
Diagram (1 mark)
Young's double-slit experiment: source, single slit S_0, double slit S_1S_2 (separation d), screen at distance L with bright and dark fringes.
Derivation of fringe spacing (2 marks)

Derivation Fringe spacing \Delta y=\lambda L/d

Aim. Find the distance between two neighbouring bright fringes.

Assumptions.
  • The screen is far away (L\gg d) and the angle \theta is small.
Path difference BC =d\sin\theta for a point P at height y above the centre O.
  1. Path difference. Draw S_1C perpendicular to S_2P. For a distant screen the rays are nearly parallel, so the path difference is BC. In the right triangle S_1S_2C the angle at S_1 equals \theta and the hypotenuse is d.
    S_2P-S_1P=BC=d\sin\theta
  2. Bright fringe. For the mth bright fringe the path difference is m whole wavelengths.
    d\sin\theta=m\lambda
  3. Position. \theta is small, so \sin\theta\approx\tan\theta=y_m/L, where y_m is the distance of the fringe from O.
    d\,\dfrac{y_m}{L}=m\lambda\;\Rightarrow\;y_m=\dfrac{m\lambda L}{d}
  4. Spacing. The next bright fringe is at y_{m+1}=(m+1)\lambda L/d. Subtract.
    \Delta y=y_{m+1}-y_m=\dfrac{(m+1)\lambda L}{d}-\dfrac{m\lambda L}{d}=\dfrac{\lambda L}{d}
Result\Delta y=\dfrac{\lambda L}{d}

The fringes are equally spaced. The spacing increases with the wavelength and the screen distance, and decreases when the slit separation increases. Dark fringes, at y=\left(m+\tfrac12\right)\lambda L/d, have the same spacing.

\Delta y=\dfrac{\lambda L}{d}

Exam focus

  1. Explain the working of Young's double-slit experiment with a labelled diagram and derive \Delta y=\lambda L/d. (long)
    Asked as Q.4 (7 marks) in both new-syllabus 2026 sets, B and D.
    Go to answer
  2. Derive the grating equation d\sin\theta=m\lambda and describe how a grating is used to find the wavelength of light. (long)
    A core learning outcome and the basis of most grating numericals and MCQs.
    Go to answer
  3. State the conditions for observing sustained interference fringes. (short)
    2026 Set D Q.2 (ii) OR, and the same as book Short question 10.
    Go to answer
  4. Explain constructive and destructive interference with their path-difference conditions. (short)
    2026 Set B Q.2 (ii) OR.
    Go to answer
  5. Define grating element and write its formula; describe the effect of more lines per unit length. (short)
    2026 Set B Q.2 (xiii) and MCQ 6, Set D Q.2 (xiii); book Short question 7.
    Go to answer
  6. How do d, L and \lambda change the fringe spacing in Young's experiment? (short)
    2026 Set B Q.2 (xiv) and Set D Q.2 (xiv); book exercise MCQ 9.
    Go to answer
  7. Numericals on \Delta y=\lambda L/d and fringe positions y=m\lambda L/d, y=\left(m+\tfrac12\right)\lambda L/d. (numerical)
    Learning outcome 3. Convert nm, \mum and mm to metres carefully: three of the book's printed answers are wrong because of such slips.
    Go to answer
  8. Numericals on the grating equation and lines per cm. (numerical)
    Learning outcome 4; a typical 3-mark numerical.
    Go to answer
  9. Explain the electron double-slit experiment and the Copenhagen and many-worlds interpretations. (long)
    Learning outcome 6; new in the 2022-23 curriculum and likely to appear as a short or long question.
    Go to answer
  10. Use the single-slit condition a\sin\theta=m\lambda in MCQs. (mcq)
    Book exercise MCQs 4 and 14 and Assignment 18.2 need it, although the text does not explain it.
    Go to answer
  11. Describe the ripple tank experiment for two-source interference. (long)
    Learning outcome 1; book Comprehensive question 1.
    Go to answer

Mark this unit as done to fill the Class 12 Physics ring.

Next, unit 19
Electric potential and capacitor

© 2026 FBISE Fold. Independent study notes, not affiliated with the Federal Board of Intermediate and Secondary Education.