Malus' law states that when plane polarised light of intensity I1 falls on an analyser, the transmitted intensity is I1 cos squared theta, where theta is the angle between the polariser and analyser axes. Unpolarised light passing a single polaroid is first reduced to half its intensity.

Diffraction

Q: What is diffraction of light?

Diffraction is the bending of light around the edges of an obstacle or aperture into the region of geometrical shadow. It is noticeable only when the obstacle or opening is comparable in size to the wavelength. It arises from interference between Huygens' secondary wavelets from different parts of the same wavefront.

Q: What is the width of the central maximum in single slit diffraction?

The first minima lie at sin theta = plus or minus lambda/a, so the central maximum has angular width 2 lambda/a and linear width 2 lambda D/a on a screen at distance D. It is twice as wide as each secondary maximum and becomes wider when the slit is narrowed or the wavelength increased.

Q: What is the difference between interference and diffraction?

Interference is the superposition of waves from two or a few coherent sources and gives many equally bright, equally wide fringes. Diffraction is the superposition of wavelets from a continuous range of points on one wavefront; its central maximum is much brighter and twice as wide as the rest, and only a few fringes are seen.

Q: What is the Rayleigh criterion?

Two point sources are just resolved when the central maximum of one diffraction image falls on the first minimum of the other. For a circular aperture of diameter D this gives the smallest resolvable angle 1.22 lambda/D. Larger apertures and shorter wavelengths give finer resolution, which is why big telescopes and electron microscopes are used.

Q: What is Brewster's law?

When unpolarised light strikes a transparent surface at the polarising angle ip, the reflected light is completely plane polarised. Brewster's law gives tan ip = mu, and at this angle the reflected and refracted rays are perpendicular. For glass of index 1.5 the polarising angle is about 56 degrees.

Q: Which diffraction and polarisation questions are asked in NEET?

NEET asks for the width of the central maximum 2 lambda D/a, the position of single slit minima, the Fresnel distance, the resolving power of telescopes and the eye, Malus' law with two or three polaroids, and Brewster's angle tan ip = mu. Remember the factor of one half for unpolarised light.

Q: How are diffraction and polarisation tested in JEE Main?

JEE Main combines single slit diffraction with YDSE (fringes inside the central maximum), asks for angular and linear widths, compares interference and diffraction, and tests Malus' law with several polaroids and Brewster's angle with Snell's law. Keeping slit width a and slit separation d apart is the key habit.

PhysicsWave OpticsFor NEET aspirants

Diffraction is the bending and spreading of light around an obstacle or through an opening whose size is comparable to its wavelength. For a single slit of width $a$ the dark directions obey $a sin θ = nλ$ , and the central maximum is $\frac{2 λ D}{a}$ wide. Diffraction also limits how finely telescopes and microscopes can resolve detail. This page also covers polarisation (Malus' and Brewster's laws), and both topics are regular one-mark questions in JEE Main and NEET.

On this page1What diffraction is2Single slit3Central maximum4Diffraction vs interference5Fresnel distance6Resolving power7Polarisation

Key Formulas - Quick Reference

Single slit minima: $a sin θ = nλ$ , $n = \pm 1, \pm 2, \dots$
Secondary maxima (approx.): $a sin θ = (n + \frac{1}{2}) λ$ ; intensity about $4.7%$ , $1.7%$ , $0.8%$ of $I_{0}$
Angular half-width of central maximum $θ \approx \frac{λ}{a}$ ; full angular width $\frac{2 λ}{a}$ ; linear width $\frac{2 λ D}{a}$
Intensity: $I = I_{0} (\frac{sin β}{β})^{2}$ , $β = \frac{π a sin θ}{λ}$
Fresnel distance: $z_{F} = \frac{a ^{2}}{λ}$ (ray optics valid for $z ≪ z_{F}$ )
Rayleigh limit (circular aperture of diameter $D$ ): $θ_{min} = \frac{1.22 λ}{D}$ ; resolving power $= \frac{1}{θ _{min}}$
Microscope: $d_{min} = \frac{1.22 λ}{2 n sin β}$ ( $n sin β$ = numerical aperture)
Malus' law: $I = I_{1} cos^{2} θ$ ; unpolarised light through a polaroid: $I = \frac{I _{0}}{2}$
Brewster's law: $tan i_{p} = μ$ ; then $i_{p} + r = 9 0^{\circ}$

1. What Is Diffraction?

★ Must learnDiffraction is the bending of waves around the edges of an obstacle or aperture, so that they spread into the region of geometrical shadow. It is noticeable only when the size of the obstacle or aperture is comparable to the wavelength.

Figure 1: Diffraction is the bending of waves around edges. By Huygens' construction only the edges of a wide gap bend the wave, but a gap about one wavelength wide acts like a single point source.

Sound ( $λ \sim 1 m$ ) bends easily around doors and walls, so we hear people we cannot see. Light ( $λ \sim 500 nm$ ) needs openings of micrometres to millimetres, which is why shadows look sharp in daily life. Look closely, though, and the edge of every shadow has faint fringes.

In Huygens' picture, every point of the unobstructed part of a wavefront is a secondary source. Diffraction is the interference of the secondary wavelets from different parts of the same wavefront. There is no real difference between the physics of interference and of diffraction; the words describe two sources versus a continuous distribution of sources.

2. Diffraction at a Single Slit

A parallel beam of monochromatic light falls normally on a slit of width $a$ . The light is viewed on a distant screen, or in the focal plane of a convex lens (Fraunhofer diffraction).

Figure 2: Rays from the two edges of a slit of width

a

, leaving at angle

θ

, differ in path by

a sin θ

. Every point of the slit is a Huygens source, so the whole slit contributes.

2.1 The central maximum

Straight ahead ( $θ = 0$ ) all secondary wavelets from the slit travel equal paths and arrive in phase, so the centre is bright: the central maximum.

2.2 Minima

Figure 3: Why

a sin θ = λ

is dark. Split the slit into halves; each point in the upper half has a partner

\frac{a}{2}

below it whose path differs by

\frac{λ}{2}

, so the halves cancel pair by pair.

Path difference between the two edges at angle $θ$ : $B N = a sin θ$ .
Let $a sin θ = λ$ . Divide the slit into two halves. A point in the upper half and the point $\frac{a}{2}$ below it (in the lower half) differ in path by $\frac{a}{2} sin θ = \frac{λ}{2}$ .
Every such pair cancels, so the two halves cancel completely: first minimum at $a sin θ = λ$ .
For $a sin θ = 2 λ$ , divide the slit into four parts; neighbouring quarters cancel: second minimum. In general:

★ Must learnMinima of a single slit:

a sin θ = nλ, n = \pm 1, \pm 2, \pm 3, \dots (never n = 0)

2.3 Secondary maxima

For $a sin θ = \frac{3 λ}{2}$ , divide the slit into three parts: two cancel and one third survives, so a weak maximum appears. Approximately,

a sin θ \approx (n + \frac{1}{2}) λ, n = 1, 2, 3, \dots

Only $\frac{1}{3}$ , then $\frac{1}{5}$ , of the slit contributes, and those parts are not in phase, so these maxima are much weaker than the central one.

Figure 4: Single slit pattern,

I = I_{0} (\frac{sin β}{β})^{2}

with

β = \frac{π a sin θ}{λ}

. The central maximum is twice as wide as the rest; secondary maxima fall to

4.7%

1.7%

0.8%

(strip brightness exaggerated so they show).

Feature	Position ( $a sin θ$ )	Relative intensity
Central maximum	0	$100%$
1st minimum	$λ$	0
1st secondary maximum	$\approx 1.43 λ$ (about $\frac{3 λ}{2}$ )	$4.7%$
2nd minimum	$2 λ$	0
2nd secondary maximum	$\approx 2.46 λ$ (about $\frac{5 λ}{2}$ )	$1.7%$

Opposite conditions! In YDSE, $d sin θ = nλ$ gives maxima. In single slit diffraction, $a sin θ = nλ$ gives minima. Check which one the question is about before writing the formula.

3. Width of the Central Maximum

Figure 5: First minima at

θ \approx \frac{λ}{a}

, i.e.

y_{1} = \frac{λ D}{a}

from the centre, so the central maximum is

\frac{2 λ D}{a}

wide (profile computed from

sinc^{2}

The central maximum extends between the first minima on either side, $sin θ = \pm \frac{λ}{a}$ . For small angles:

Quantity	Formula
Angular half-width (angle of first minimum)	$θ_{1} = \frac{λ}{a}$
Angular width of central maximum	$2 θ_{1} = \frac{2 λ}{a}$
Distance of $n$ th minimum from centre	$y_{n} = \frac{nλ D}{a}$
Linear width of central maximum	$\frac{2 λ D}{a}$
Width of each secondary maximum	$\frac{λ D}{a}$ (half the central one)

Figure 6: Narrow slit, wide pattern. Doubling

a

halves the angular width

\frac{2 λ}{a}

of the central maximum; a longer wavelength widens it.

Exam Trick

Squeeze the slit, spread the light. Width of the central maximum $\propto \frac{λ}{a}$ . Halve the slit: pattern twice as wide. Use red instead of blue: wider. Put the set-up in water: $λ \to \frac{λ}{μ}$ , narrower. If $a < λ$ there is no minimum at all ( $sin θ$ would exceed 1): light spreads over the whole screen.

Key idea

Narrower slit or longer wavelength means a wider central maximum: angular width

\frac{2 λ}{a}

, linear width

\frac{2 λ D}{a}

4. Interference and Diffraction Compared

Interference (YDSE)	Diffraction (single slit)
Superposition of waves from two (or a few) coherent sources	Superposition of wavelets from a continuous set of points on one wavefront
Maxima at $d sin θ = nλ$	Minima at $a sin θ = nλ$
All bright fringes (nearly) equally bright	Central maximum much brighter; secondary maxima fade quickly
All fringes have equal width $\frac{λ D}{d}$	Central maximum twice as wide ( $\frac{2 λ D}{a}$ ) as the others
Many fringes visible	Only a few fringes visible
Minima are perfectly dark (equal slits)	Minima are dark, but the pattern is dominated by the central peak

Figure 7: A real YDSE has slits of finite width

a

, so its

cos^{2}

fringes sit inside the single-slit envelope. Here

d = 4 a

: 7 bright fringes fit in the central envelope and order

\pm 4

is missing.

JEE Advanced

Double slit with finite slit width. The intensity is $I = I_{0} cos^{2} (\frac{π d sin θ}{λ}) (\frac{sin β}{β})^{2}$ : YDSE fringes inside a single-slit envelope. The number of bright fringes inside the central envelope is about $\frac{2 d}{a}$ ( $\frac{2 d}{a} - 1$ if an interference maximum falls exactly on the envelope minimum, which is then a missing order). Orders with $\frac{n}{d} = \frac{m}{a}$ are missing.

Quick Recall: tap to check

Where is the first minimum of a single slit of width

a

a sin θ = λ

Angular width of the central maximum?

\frac{2 λ}{a}

(from

- \frac{λ}{a}

+ \frac{λ}{a}

Why is the central maximum twice as wide as the others?

It spans two first minima,

\pm \frac{λ}{a}

; every other maximum lies between minima only

\frac{λ}{a}

apart.

A double slit has

d = 5 a

. Which order is missing first?

The 5th, where

d sin θ = 5 λ

meets the first diffraction minimum.

5. Fresnel Distance: When Ray Optics Is Valid

A beam from an aperture of width $a$ spreads by diffraction at an angle of about $\frac{λ}{a}$ . After travelling a distance $z$ , the spread is about $\frac{z λ}{a}$ . This becomes comparable to the aperture itself when $\frac{z λ}{a} = a$ :

★ Must learnFresnel distance

z_{F} = \frac{a ^{2}}{λ}

For

z ≪ z_{F}

the beam keeps its width and ray optics works; for

z ≫ z_{F}

diffraction dominates. Ray optics is the limit

λ \to 0

(

z_{F} \to \infty

Figure 8: Up to the Fresnel distance

z_{F} = \frac{a ^{2}}{λ}

the diffraction spread

\frac{z λ}{a}

is smaller than the aperture, so ray optics works; beyond it diffraction dominates.

6. Resolving Power of Optical Instruments

Because of diffraction at the lens aperture, even a perfect lens images a point source as a small bright disc surrounded by faint rings (the Airy pattern), not as a point. Two nearby points whose discs overlap too much cannot be seen as separate.

★ Must learnRayleigh criterion: two point sources are just resolved when the central maximum of one image falls on the first minimum of the other. For a circular aperture of diameter

D

θ_{min} = \frac{1.22 λ}{D}

θ_{min}

is the limit of resolution; its reciprocal is the resolving power.

Figure 9: Rayleigh criterion (computed Airy patterns). Two images are just resolved when the central maximum of one falls on the first minimum of the other:

θ_{min} = \frac{1.22 λ}{D}

Instrument	Limit of resolution	Resolving power	Improves with
Telescope (objective diameter $D$ )	$Δ θ = \frac{1.22 λ}{D}$	$\frac{D}{1.22 λ}$	Larger $D$ , shorter $λ$
Human eye (pupil about $3 mm$ )	about $2 \times 1 0^{- 4} rad$ (about $1^{'}$ )		Brighter light (wider pupil)
Microscope (half-angle $β$ , medium $n$ )	$d_{min} = \frac{1.22 λ}{2 n sin β}$	$\frac{2 n sin β}{1.22 λ}$	Oil immersion (larger $n$ ), shorter $λ$ (UV, electrons)

Exam Trick

Bigger and bluer is sharper. Resolving power $\propto \frac{D}{λ}$ . That is why astronomers build huge telescopes, why microscopes use oil immersion ( $n sin β$ = numerical aperture, up to about 1.5) and why electron microscopes, with $λ \sim 1 0^{- 11} m$ , see far smaller detail.

Telescope

Resolves angles between distant objects: $Δ θ = \frac{1.22 λ}{D}$ . Improve it with a bigger objective $D$ . Magnification does not help if the images are not resolved.

Microscope

Resolves distances between nearby points: $d_{min} = \frac{1.22 λ}{2 n sin β}$ . Improve it with oil immersion (larger $n$ ) or a shorter wavelength.

Key idea

Resolution improves with a bigger aperture and a shorter wavelength:

θ_{min} = \frac{1.22 λ}{D}

7. Polarisation

Light is a transverse wave: its electric field $E$ vibrates perpendicular to the direction of travel. Polarisation is the effect that shows this, and longitudinal waves such as sound cannot be polarised.

★ Must learnPolarisation: restricting the vibrations of

E

to a single direction perpendicular to the ray. Light with

E

in one direction only is plane (linearly) polarised; the plane containing

E

and the ray is the plane of vibration.

Figure 10: Polarisation symbols. End-on, unpolarised light has

E

in every direction perpendicular to the ray; plane polarised light has one direction only. Side on, arrows show

E

in the page and dots show

E

perpendicular to it.

7.1 Polaroids and Malus' law

A polaroid is a sheet that transmits only the component of $E$ along its transmission axis. Unpolarised light through one polaroid becomes plane polarised with half the intensity (the average of $cos^{2} θ$ over all directions is $\frac{1}{2}$ ). A second polaroid, the analyser, at angle $θ$ to the first passes only the component $E cos θ$ :

★ Must learnMalus' law:

I = I_{1} cos^{2} θ

, where

I_{1}

is the intensity of the plane polarised light falling on the analyser and

θ

the angle between the transmission axes.

Figure 11: Malus' law. A polariser halves unpolarised light (

I_{1} = \frac{I _{0}}{2}

); an analyser at angle

θ

passes

I_{1} cos^{2} θ

. Rotating the analyser gives two maxima and two zeros per turn only for plane polarised light.

7.2 How to test a beam of light

Look at the beam through a polaroid (analyser) and rotate it once:

What you see	The beam is
No change in intensity	Unpolarised (or circularly polarised)
Intensity varies but never becomes zero	Partially polarised
Intensity varies and falls to zero twice per rotation	Plane polarised

7.3 Polarisation by reflection: Brewster's law

Figure 12: At the polarising angle (

i_{p} = 56. 3^{\circ}

for

μ = 1.5

) the reflected light is plane polarised and makes

9 0^{\circ}

with the refracted ray (

r = 33. 7^{\circ}

When unpolarised light strikes a transparent surface, the reflected light is partly polarised. At one special angle of incidence, the polarising (Brewster) angle $i_{p}$ , the reflected light is completely plane polarised, with $E$ perpendicular to the plane of incidence. At this angle the reflected and refracted rays are perpendicular.

Condition: $i_{p} + r = 9 0^{\circ}$ , so $r = 9 0^{\circ} - i_{p}$ .
Snell's law: $μ = \frac{sin i _{p}}{sin r} = \frac{sin i _{p}}{sin ( 9 0 ^{\circ} - i _{p} )} = \frac{sin i _{p}}{cos i _{p}}$ .
Hence
$tan i_{p} = μ (Brewster’s law)$
For glass ( $μ = 1.5$ ), $i_{p} \approx 56. 3^{\circ}$ ; for water ( $1.33$ ), $i_{p} \approx 53. 1^{\circ}$ .

7.4 Polarisation by scattering

Sunlight scattered by air molecules is polarised: light scattered at $9 0^{\circ}$ to the incoming sunlight is almost plane polarised. Look at the blue sky at right angles to the Sun through a rotating polaroid and its brightness changes. Bees use this polarisation of skylight to navigate.

7.5 Uses of polaroids

Sunglasses and camera filters: they cut the glare of light reflected from roads, water and glass, which is partly horizontally polarised.
3D films: the two pictures are projected with perpendicular polarisations and each eye's filter passes only one.
Liquid crystal displays (LCD screens, calculators) work by switching polarisation.
Car headlights and windshields with crossed polaroids to reduce night-time glare from oncoming cars.
Photoelastic stress analysis: stressed plastic between crossed polaroids shows coloured stress patterns.

Key idea

Only transverse waves can be polarised. A polaroid halves unpolarised light; after that, Malus' law

I = I_{1} cos^{2} θ

takes over.

Quick Recall: tap to check

Unpolarised light

I_{0}

passes two polaroids at

3 0^{\circ}

. Output?

\frac{I _{0}}{2} cos^{2} 3 0^{\circ} = \frac{3 I _{0}}{8}

Brewster angle for glass of

μ = 1.5

tan^{- 1} 1.5 = 56. 3^{\circ}

Can sound in air be polarised?

No: it is a longitudinal wave, so there is no sideways vibration to select.

Resolving power of a telescope is proportional to?

\frac{D}{λ}

: aperture over wavelength.

Figure 13: Pick the formula from the question: slit directions, central width

\frac{2 λ D}{a}

, resolution

1.22 \frac{λ}{D}

, Fresnel distance

\frac{a ^{2}}{λ}

, or Malus and Brewster for polarised light.

Figure 14: Revision map: single slit, central maximum

\frac{2 λ D}{a}

, comparison with interference, resolution

1.22 \frac{λ}{D}

, Fresnel distance and polarisation.

8. Solved Examples

Solved Example 1

Light of wavelength

600 nm

falls normally on a slit

0.2 mm

wide. The pattern is seen on a screen

2 m

away. Find the angular half-width and the linear width of the central maximum.

Solution:

$θ_{1} = \frac{λ}{a} = \frac{6 \times 1 0 ^{- 7}}{2 \times 1 0 ^{- 4}} = 3 \times 1 0^{- 3} rad$ .

Linear width $= \frac{2 λ D}{a} = 2 \times 3 \times 1 0^{- 3} \times 2 = 1.2 \times 1 0^{- 2} m$ .

Answer: $3 \times 1 0^{- 3} rad$ (about $0.1 7^{\circ}$ ); $12 mm$ .

Solved Example 2

The first minimum of a single slit pattern for light of

500 nm

is at

3 0^{\circ}

. Find the slit width and the direction of the first secondary maximum.

Solution:

$a sin 3 0^{\circ} = λ \Rightarrow a = \frac{500 nm}{0.5} = 1000 nm = 1 μ m$ .

First secondary maximum: $a sin θ \approx \frac{3 λ}{2} \Rightarrow sin θ = \frac{3}{2} \times \frac{1}{2} = 0.75$ , so $θ \approx 48. 6^{\circ}$ .

Answer: $a = 1 μ m$ ; about $48. 6^{\circ}$ . (Small-angle formulas would fail here: $a$ is only $2 λ$ .)

Solved Example 3

In a double slit experiment (

d = 1 mm

λ = 500 nm

), what should the width of each slit be so that 10 maxima of the double slit pattern lie within the central maximum of the single slit pattern?

Solution:

Angular width of the central diffraction maximum: $\frac{2 λ}{a}$ . Angular spacing of the interference maxima: $\frac{λ}{d}$ .

For 10 fringes inside: $10 \times \frac{λ}{d} = \frac{2 λ}{a} \Rightarrow a = \frac{d}{5} = 0.2 mm$ .

Answer: $0.2 mm$ (the wavelength cancels out).

Solved Example 4

For what distance is ray optics a good approximation for an aperture

3 mm

wide and light of wavelength

500 nm

Solution:

$z_{F} = \frac{a ^{2}}{λ} = \frac{( 3 \times 1 0 ^{- 3} ) ^{2}}{5 \times 1 0 ^{- 7}} = \frac{9 \times 1 0 ^{- 6}}{5 \times 1 0 ^{- 7}} = 18 m$ .

Answer: up to about $18 m$ ; beyond that the beam spreads noticeably by diffraction.

Solved Example 5

The pupil of the eye is about

3 mm

across. Taking

λ = 500 nm

, find the limit of resolution of the eye. Up to what distance could the two headlights of a car (

1.5 m

apart) be seen as separate, ignoring the atmosphere?

Solution:

$θ_{min} = \frac{1.22 λ}{D} = \frac{1.22 \times 5 \times 1 0 ^{- 7}}{3 \times 1 0 ^{- 3}} = 2.03 \times 1 0^{- 4} rad$ (about $0. 7^{'}$ ).

The lights are resolved while $\frac{1.5}{L} \geq θ_{min}$ : $L \leq \frac{1.5}{2.03 \times 1 0 ^{- 4}} \approx 7.4 \times 1 0^{3} m$ .

Answer: about $2 \times 1 0^{- 4} rad$ ; up to about $7.4 km$ .

Solved Example 6

A space telescope has a mirror of diameter

2.4 m

. Find its angular limit of resolution for light of

550 nm

. How does it compare with the eye?

Solution:

$θ_{min} = \frac{1.22 \times 5.5 \times 1 0 ^{- 7}}{2.4} = 2.8 \times 1 0^{- 7} rad$ .

Compared with the eye ( $2 \times 1 0^{- 4} rad$ ) it is about $700$ times finer, because $θ_{min} \propto \frac{1}{D}$ .

Answer: $2.8 \times 1 0^{- 7} rad$ .

Solved Example 7

An oil-immersion microscope objective has numerical aperture

n sin β = 1.4

. Find the smallest separation it can resolve with light of

550 nm

Solution:

$d_{min} = \frac{1.22 λ}{2 n sin β} = \frac{1.22 \times 550 nm}{2 \times 1.4} = 240 nm$ .

Answer: about $0.24 μ m$ . Finer detail needs shorter wavelengths (ultraviolet, or electrons).

Solved Example 8

Unpolarised light of intensity

I_{0}

passes through two polaroids whose axes are at

6 0^{\circ}

. Find the transmitted intensity.

Solution:

First polaroid: $I_{1} = \frac{I _{0}}{2}$ . Second (Malus): $I_{2} = I_{1} cos^{2} 6 0^{\circ} = \frac{I _{0}}{2} \times \frac{1}{4}$ .

Answer: $\frac{I _{0}}{8}$ .

Solved Example 9

Two polaroids are crossed (axes at

9 0^{\circ}

), so no light passes. A third polaroid is placed between them with its axis at

4 5^{\circ}

to each. Find the intensity that now emerges if unpolarised light of intensity

I_{0}

is incident.

Solution:

After the first: $\frac{I _{0}}{2}$ . After the middle one ( $4 5^{\circ}$ ): $\frac{I _{0}}{2} cos^{2} 4 5^{\circ} = \frac{I _{0}}{4}$ . After the last ( $4 5^{\circ}$ again): $\frac{I _{0}}{4} cos^{2} 4 5^{\circ} = \frac{I _{0}}{8}$ .

Answer: $\frac{I _{0}}{8}$ . Inserting a polaroid between crossed polaroids lets light through, because it rotates the plane of polarisation in two steps.

Solved Example 10

Find the polarising angle for glass of refractive index

3

, and the angle of refraction at that incidence.

Solution:

$tan i_{p} = 3 \Rightarrow i_{p} = 6 0^{\circ}$ . Then $r = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ .

Check with Snell's law: $\frac{sin 6 0 ^{\circ}}{sin 3 0 ^{\circ}} = \frac{3 /2}{1/2} = 3$ .

Answer: $i_{p} = 6 0^{\circ}$ , $r = 3 0^{\circ}$ .

Solved Example 11

A beam is viewed through a rotating polaroid. Its intensity changes between a maximum and a non-zero minimum. The beam is
(A) unpolarised
(B) plane polarised
(C) partially polarised
(D) monochromatic

Solution:

Unpolarised light shows no change; plane polarised light falls to zero twice per rotation. A change without zero means one direction of $E$ is stronger but not the only one.

Answer: (C).

Solved Example 12

Unpolarised light falls on a glass plate (

μ = 1.5

). The reflected light is completely plane polarised when the angle of incidence is closest to
(A)

33. 7^{\circ}

(B)

4 5^{\circ}

(C)

56. 3^{\circ}

(D)

9 0^{\circ}

Solution:

Brewster's law: $tan i_{p} = μ = 1.5 \Rightarrow i_{p} = tan^{- 1} 1.5 = 56. 3^{\circ}$ . Then the refracted ray makes $r = 9 0^{\circ} - 56. 3^{\circ} = 33. 7^{\circ}$ with the normal.

Answer: (C). Option (A), $33. 7^{\circ}$ , is the angle of refraction, a common trap.

Practice Questions

Light of $500 nm$ falls on a slit $0.1 mm$ wide; the screen is $1 m$ away. Find the width of the central maximum.Answer: $\frac{2 λ D}{a} = 10 mm$ .
A slit has width $a = 2 λ$ . Find the full angular width of the central maximum.Answer: $sin θ = \frac{1}{2}$ , so $θ = 3 0^{\circ}$ each side: $6 0^{\circ}$ .
What fraction of the central intensity is the first secondary maximum of a single slit?Answer: About $\frac{4}{9 π ^{2}} \approx 4.5%$ (exact value $4.7%$ ).
Unpolarised light passes through three polaroids, each axis turned $3 0^{\circ}$ from the previous one. Find the final intensity.Answer: $\frac{I _{0}}{2} cos^{2} 3 0^{\circ} cos^{2} 3 0^{\circ} = \frac{9 I _{0}}{32}$ .
Find the polarising angle for water ( $μ = 1.33$ ).Answer: $tan^{- 1} (1.33) \approx 53. 1^{\circ}$ .
Find the angular limit of resolution of a telescope with a $1 m$ objective for $λ = 600 nm$ .Answer: $\frac{1.22 \times 6 \times 1 0 ^{- 7}}{1} = 7.3 \times 1 0^{- 7} rad$ .

Common Mistakes to Avoid

Watch out

Using $a sin θ = nλ$ for single-slit maxima. It gives the minima; the maxima are near $(n + \frac{1}{2}) λ$ .
Taking the width of the central maximum as $\frac{λ D}{a}$ . That is the half-width; the full width is $\frac{2 λ D}{a}$ .
Mixing up $d$ (slit separation, YDSE) and $a$ (slit width, diffraction) in combined problems.
Thinking a narrower slit gives a narrower pattern. The pattern width varies as 1/a: narrower slit, wider pattern.
Forgetting the factor $\frac{1}{2}$ when unpolarised light passes the first polaroid, or writing $cos θ$ instead of $cos^{2} θ$ in Malus' law afterwards (amplitude goes as $cos θ$ , intensity as $cos^{2} θ$ ).
Using $sin i_{p} = μ$ or $cos i_{p}$ : Brewster's law is $tan i_{p} = μ$ , with $i_{p} + r = 9 0^{\circ}$ .
Confusing limit of resolution (smaller is better, $\frac{1.22 λ}{D}$ ) with resolving power (larger is better, $\frac{D}{1.22 λ}$ ).
Saying sound can be polarised. Only transverse waves can be polarised; sound in air is longitudinal.

Frequently Asked Questions

What is diffraction of light?

Diffraction is the bending of light around the edges of an obstacle or aperture into the region of geometrical shadow. It is noticeable only when the obstacle or opening is comparable in size to the wavelength. It arises from interference between Huygens' secondary wavelets from different parts of the same wavefront.

What is the width of the central maximum in single slit diffraction?

The first minima lie at $sin θ = \pm λ / a$ , so the central maximum has angular width $2 λ / a$ and linear width $2 λ D / a$ on a screen at distance $D$ . It is twice as wide as each secondary maximum and becomes wider when the slit is narrowed or the wavelength increased.

What is the difference between interference and diffraction?

Interference is the superposition of waves from two or a few coherent sources and gives many equally bright, equally wide fringes. Diffraction is the superposition of wavelets from a continuous range of points on one wavefront; its central maximum is much brighter and twice as wide as the rest, and only a few fringes are seen.

What is the Rayleigh criterion?

Two point sources are just resolved when the central maximum of one diffraction image falls on the first minimum of the other. For a circular aperture of diameter $D$ this gives the smallest resolvable angle $1.22 λ / D$ . Larger apertures and shorter wavelengths give finer resolution, which is why big telescopes and electron microscopes are used.

What is Malus' law?

Malus' law states that when plane polarised light of intensity $I_{1}$ falls on an analyser, the transmitted intensity is $I_{1} cos^{2} θ$ , where $θ$ is the angle between the polariser and analyser axes. Unpolarised light passing a single polaroid is first reduced to half its intensity.

What is Brewster's law?

When unpolarised light strikes a transparent surface at the polarising angle $i_{p}$ , the reflected light is completely plane polarised. Brewster's law gives $tan i_{p} = μ$ , and at this angle the reflected and refracted rays are perpendicular. For glass of index 1.5 the polarising angle is about 56 degrees.

Which diffraction and polarisation questions are asked in NEET?

NEET asks for the width of the central maximum $2 λ D / a$ , the position of single slit minima, the Fresnel distance, the resolving power of telescopes and the eye, Malus' law with two or three polaroids, and Brewster's angle $tan i_{p} = μ$ . Remember the factor of one half for unpolarised light.

How are diffraction and polarisation tested in JEE Main?

JEE Main combines single slit diffraction with YDSE (fringes inside the central maximum), asks for angular and linear widths, compares interference and diffraction, and tests Malus' law with several polaroids and Brewster's angle with Snell's law. Keeping slit width a and slit separation d apart is the key habit.

Previous year questions on Diffraction

10 questions from past papers, each with a step-by-step solution.

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