Wave Optics: theory | Physics Circle

CONCEPT

Wave Theory, Wavefronts & Doppler Effect

In one line
Light travels as a wave. Every point on a wavefront starts new wavelets, and their envelope is the next wavefront.

THEORIES

7.1.1 Newton and Huygens

Newton: corpuscles	Huygens: waves
tiny particles from a luminous body	disturbance travelling as a wave
colour = size of corpuscle	colour = wavelength
speed greater in a denser medium (wrong)	speed smaller in a denser medium (correct)
fails: interference, diffraction, polarisation	explains: interference, diffraction, reflection, refraction

Huygens first took light waves as longitudinal; polarisation showed they are transverse (Fresnel).
The wave theory cannot explain the photoelectric effect, Compton effect or origin of spectra.

WAVEFRONT

7.1.2 What a wavefront is

A wavefront is the locus of all particles vibrating in the same phase.
A ray is always perpendicular to the wavefront.
Phase difference between two points on one wavefront is zero.

Spherical, cylindrical and plane wavefronts

Wavefront	Source	I	A
spherical	point	∝ 1r²	∝ 1r
cylindrical	line	∝ 1r	∝ 1√r
plane	very far	constant	constant

Parallel rays: plane. Converging rays: spherical, radius decreasing

A source at the focus of a convex lens gives a plane wavefront.
Diverging rays: spherical wavefront of increasing radius.

HUYGENS

7.1.3 Huygens’ principle

Secondary wavelets and the new wavefront

Every point on a wavefront is a source of secondary wavelets.
They spread with the speed of light in that medium.
The surface touching them in the forward direction is the new wavefront.

REFLECTION, REFRACTION

7.1.4 Wavefronts at a boundary

Reflection: BC = AD

Reflection: speed is unchanged, so BC = AD and ∠i = ∠r.
Widths of incident and reflected wavefronts are equal.

Refraction: light goes BC and AD in the same time

BCAD = v₁v₂ = sin isin r = μ₂μ₁

width of refracted wavefrontwidth of incident wavefront = cos rcos i

LIGHT IN A MEDIUM

7.1.5 What changes and what does not

v = cμ · λ′ = λμ · ν′ = ν

Frequency (and photon energy) does not change.
Speed, wavelength and momentum of the photon change.
μλ is constant for the same light.
Ray optics is valid when the size of the obstacle is much larger than λ.

DOPPLER EFFECT

7.1.6 Doppler effect of light

Δλλ = Δνν = vc (v ≪ c)

Source approaching: λ decreases, shift towards violet (blue shift).
Source receding: λ increases, shift towards red (red shift).
Used to find speeds of stars and galaxies.

Practise PYQs on this topicWave Theory, Wavefronts & Doppler Effect · 29 questions with solutions

CONCEPT

Superposition, Coherence & Intensity

In one line
Two coherent waves add at every point. Intensity is largest where they meet in phase and smallest where they meet in opposite phase.

SUPERPOSITION

7.2.1 Adding two waves

The resultant (thick line) is the sum of the two waves

Resultant displacement is the vector sum: y = y₁ + y₂.
Take y₁ = a₁ sin ωt and y₂ = a₂ sin(ωt + φ).

A² = a₁² + a₂² + 2a₁a₂ cos φ

Equal amplitudes a: A = 2a cos φ2.
Intensity ∝ (amplitude)².
Intensity ∝ slit width (if amplitude ∝ width, intensity ∝ width²).

I = I₁ + I₂ + 2√I₁I₂ cos φ

PHASE AND PATH

7.2.2 Phase and path difference

φ = 2πλ Δ · time difference = T2π φ

COHERENCE

7.2.3 Coherent sources

Coherent sources: same frequency and a constant phase difference.
Two independent lamps are never coherent.
They are made from one source, by two methods:

Division of wavefront (YDSE, biprism) and of amplitude (thin films)

Division of wavefront: virtual or imaginary sources. Examples: YDSE, Fresnel’s biprism, Lloyd’s mirror.
Division of amplitude: real sources, by partial reflection. Examples: thin films, Newton’s rings.

CONDITIONS

7.2.4 Conditions for sustained interference

Sources must be coherent: same frequency, constant phase difference.
Red and blue filters on the two slits: no fringes.
One slit covered: no interference fringes.
Light should be monochromatic and sources close to each other.
Equal amplitudes give the best contrast: I_max = 4I₀, I_min = 0.
Energy is not lost, only redistributed: conservation of energy.
Interference occurs for both light and sound waves.

MAXIMA, MINIMA

7.2.5 Bright and dark fringes

	Bright (constructive)	Dark (destructive)
phase φ	2nπ	(2n − 1)π
path Δ	nλ	(2n − 1)λ2
amplitude	a₁ + a₂	a₁ − a₂
intensity	(√I₁ + √I₂)²	(√I₁ − √I₂)²
I₁ = I₂ = I₀	4I₀	0

I_maxI_min = (a₁ + a₂a₁ − a₂)² = (√I₁ + √I₂√I₁ − √I₂)²

Given I_max : I_min, first take the root, then find a₁ : a₂.
One slit made wider: maxima and minima both get brighter; dark fringes are not fully dark.

INTENSITY AT A POINT

7.2.6 Intensity at a point

I = 4I₀ cos²φ2 = I_max cos²φ2

Path Δ	Phase φ	II_max
0	0	1
λ6	60°	34
λ4	90°	12
λ3	120°	14
λ2	180°	0
λ	360°	1

I₀ is the intensity of one slit; I_max = 4I₀.
For a position x on the screen: Δ = xdD, then use the table.
For intensity given: find φ, then Δ, then x or θ.

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Coherence time and coherence length

A light wave stays sinusoidal only for a short time, the coherence time τ_c.
Atoms: τ_c ≈ 10⁻⁹ to 10⁻¹⁰ s. Laser: τ_c ≈ 10⁻⁵ s.

coherence length L = cτ_c

spectral width Δλ ≈ λ²L

Average intensity and fringe visibility

I_av = I_max + I_min2 = I₁ + I₂

V = I_max − I_minI_max + I_min = 2√I₁I₂I₁ + I₂

V = 1 when I_min = 0: the best fringes.
V = 0 when I_max = I_min: no fringes.

Practise PYQs on this topicSuperposition, Coherence & Intensity · 103 questions with solutions

CONCEPT

YDSE: Fringe Width & Fringe Positions

In one line
Two narrow slits act as coherent sources and give equally spaced bright and dark fringes on the screen.

THE SET-UP

7.3.1 Young’s double slit experiment

S₁ and S₂ act as coherent sources

Centre of the screen: Δ = 0, so the central fringe is always bright.
Fringes are straight lines parallel to the slits.
All fringes have equal width.

PATH DIFFERENCE

7.3.2 At a point P on the screen

S₂M is the path difference

Δ = d sin θ = xdD

Bright: Δ = nλ, n = 0, ±1, ±2 …
Dark: Δ = (2n − 1)λ2, n = 1, 2, 3 …
Δ = 6λ means the 6th bright band.
Δ = 572λ: 2n − 1 = 57, the 29th dark band.
Dark band between 4th and 5th bright: Δ = 4.5λ.

FRINGE WIDTH

7.3.3 Fringe width β

β = λDd

β is the distance between two neighbouring bright (or dark) fringes.
β ∝ λ: red gives wider fringes than blue.
β ∝ D: move the screen away, fringes widen.
β ∝ 1d: bring the slits closer, fringes widen.
For the same β, keep λDd the same.

Moving the screen: Δβ = λ ΔDd.
A graph of β against D is a straight line through the origin, slope λd.

POSITIONS

7.3.4 Positions of the fringes

nᵗʰ bright: x = nβ = nλDd

nᵗʰ dark: x = (n − 12)β = (2n − 1)λD2d

Between	Same side	Opposite sides
nᵗʰ and mᵗʰ bright	(n − m)β	(n + m)β
nᵗʰ bright, mᵗʰ dark	n − (m − 12) times β	n + (m − 12) times β

Example: 5th bright and 3rd dark on one side: 5β − 2.5β = 2.5β.
The nth dark fringe lies between the (n − 1)th and nth bright fringes.

A FIXED REGION

7.3.5 Number of fringes in a region

n₁λ₁ = n₂λ₂

In one region of the screen, a smaller λ gives more fringes.

ANGULAR WIDTH

7.3.6 Angular width θ

θ = βD = λd

Angular width does not depend on D.
Moving the screen changes β but not θ.
Angle between the two wave trains reaching the screen: λβ.

WHITE LIGHT

7.3.7 YDSE with white light

Centre: all colours have Δ = 0, so the central fringe is white.
On either side a few coloured fringes, violet nearest and red farthest.
Beyond that: uniform illumination.

Practise PYQs on this topicYDSE: Fringe Width & Fringe Positions · 78 questions with solutions

CONCEPT

YDSE: Film & Medium

In one line
A transparent film adds optical path and shifts the whole pattern towards it. A medium shrinks the wavelength and the fringes.

OPTICAL PATH

7.4.1 Path measured in vacuum

optical path = μ × geometrical path

A film of thickness t in air adds an extra path of (μ − 1)t.
Two paths in two media: Δ = μ₁L₁ − μ₂L₂.

φ = 2πλ(μ₁L₁ − μ₂L₂)

λ here is the wavelength in vacuum.

FILM IN ONE BEAM

7.4.2 Shift of the pattern

The pattern shifts towards the slit covered by the film

Whole pattern shifts towards the film side.
Fringe width does not change.
Number of fringes seen does not change.

shift = Dd(μ − 1)t = βλ(μ − 1)t

HOW MANY FRINGES

7.4.3 Number of fringes shifted

(μ − 1)t = nλ

Number of fringes shifted: n = (μ − 1)tλ.
Same plate, different μ: shift ∝ (μ − 1).
Intensity at the old centre unchanged: least t = λμ − 1.
Old centre becomes dark: least t = λ2(μ − 1).
Equal films on both slits: extra path = (μ₂ − μ₁)t.
Point changes from dark (order m) to bright (order n): use the difference in Δ.

WHOLE SET-UP IN A MEDIUM

7.4.4 Whole set-up in a medium

λ′ = λμ · β′ = βμ · θ′ = θμ

Fringe width and angular width both decrease.
In water (μ = 43): β′ = 34β.
Air to vacuum: fringes get very slightly wider.
mᵗʰ fringe in the medium at the place of nᵗʰ fringe in air: compare positions, ratio gives μ.

Practise PYQs on this topicYDSE: Film, Medium & Two Wavelengths · 73 questions with solutions

CONCEPT

YDSE: Two Wavelengths & Opposite a Slit

In one line
For two colours, fringes coincide where their positions match. A point opposite a slit has a path difference of d²2D.

SAME ORDER

7.5.1 Same order of two wavelengths

Δx = nDd(λ₂ − λ₁)

Same order n of two wavelengths lies at different places.
Larger wavelength lies farther from the centre.

BRIGHT ON BRIGHT

7.5.2 Bright fringe on bright fringe

n₁λ₁ = n₂λ₂

Reduce n₁n₂ = λ₂λ₁ to its lowest whole numbers.
Example: 480 nm and 600 nm: n₁ : n₂ = 5 : 4.
Least distance from the centre: x = n₁β₁ = n₂β₂.
The central fringe always coincides for all colours.

BRIGHT ON DARK

7.5.3 Bright fringe on dark fringe

nλ₁ = (2m − 1)λ₂2

nᵗʰ bright of λ₁ on mᵗʰ dark of λ₂.
Then λ₁λ₂ = 2m − 12n.

POINT OPPOSITE A SLIT

7.5.4 Point opposite a slit

P lies straight in front of S₁

Δ = xdD = d²2D

This comes from S₂P − S₁P when D ≫ d.

dark at P: λ = d²(2n − 1)D

First dark: λ = d²D. Second: d²3D. Third: d²5D.
Fourth: d²7D. Fifth: d²9D.
Possible wavelengths are inversely proportional to D, 3D, 5D …
These are the missing wavelengths at P in white light.

bright at P: λ = d²2nD

Intensity at P when d and D are given: find Δ = d²2D, then φ, then I = I_max cos²φ2.

JEE · NEET

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Number of maxima on the screen

d sin θ = nλ and sin θ cannot exceed 1.
So n ≤ dλ.
If dλ = 4.6, take n = 0, ±1, ±2, ±3, ±4: total 9 maxima.

Practise PYQs on this topicYDSE: Film, Medium & Two Wavelengths · 73 questions with solutions

CONCEPT

Biprism & Thin Films

In one line
A biprism splits one slit into two virtual coherent sources, so all the YDSE formulas apply. Thin films give colours by interference of reflected light.

BIPRISM

7.6.1 Fresnel’s biprism

S₁ and S₂ are virtual images of the slit S

Two thin prisms joined base to base.
Acute angles about ½°, obtuse angle about 179°.
It forms two virtual coherent sources S₁ and S₂.
Fringes are seen through an eyepiece with a micrometer.
D is the distance from slit to eyepiece.

β = λDd, D = a + b

All YDSE results apply: positions, coincidences, intensity, film shift.
β against D is a straight line through the origin.
For constant d, βD is the same for all readings.
n₁λ₁ = n₂λ₂ in the same region.
Dark band opposite a slit: λ = d²(2n − 1)D.
Eyepiece moved x across n fringes: β = xn.

FINDING d

7.6.2 Finding the slit separation d

d = 2a(μ − 1)α

α = angle of the prism, a = slit to biprism distance.
A convex lens between biprism and eyepiece gives sharp images at two positions.
If the separations of the images are d₁ and d₂:

d = √d₁d₂

λ = βdD

THIN FILMS

7.6.3 Thin film interference

Rays reflected from the two faces interfere

Colours of soap bubbles and oil on water are due to interference.
Reflection from a denser medium adds a phase π.

Light	Maximum	Minimum
reflected	2μt cos r = (2n − 1)λ2	2μt cos r = nλ
transmitted	2μt cos r = nλ	2μt cos r = (2n − 1)λ2

Normal incidence: cos r = 1.
Reflected and transmitted patterns are complementary.

PLATE

7.6.4 Plate with partial reflection

Reflected fraction r at the top: I₁ = rI₀, and (1 − r)I₀ enters.
Lower face reflects a fraction of what enters.
Then use I_max : I_min with these intensities.

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Lloyd’s mirror

S₁ and its image S₂ act as coherent sources

Light reflected at grazing incidence from a denser medium gains a phase π.
Fringe width is the same as in YDSE, but maxima and minima are interchanged.
At the mirror edge the fringe is dark.

Newton’s rings

Air film between lens and plate gives circular fringes

Central fringe in reflected light is a dark spot.

dark ring: r_n = √nλR

bright ring: r_n = √(n + 12)λR

liquid of index μ: r_n = √nλRμ

λ = D²_n+p − D²_n4pR

Practise PYQs on this topicBiprism, Thin Films & Other Devices · 30 questions with solutions

CONCEPT

Single Slit: Minima & Central Maximum

In one line
Light bends at a narrow slit. The pattern has a wide, bright central maximum, with dark minima where the slit divides into pairs of opposite wavelets.

DIFFRACTION

7.7.1 Diffraction of light

Bending of light around an obstacle or aperture of size comparable to λ.
Larger λ gives more diffraction.
Sunlight bending at a mountain’s edge gives a luminous border.
It is the superposition of wavelets from one wavefront.

Type	Source and screen
Fresnel	at a finite distance
Fraunhofer	effectively at infinity (uses lenses)

SINGLE SLIT

7.7.2 Fraunhofer diffraction at a single slit

Parallel light on a slit of width b; P is in the focal plane

path difference between edges = b sin θ

MINIMA

7.7.3 Minima

b sin θ = nλ, n = 1, 2, 3 …

x_n = nλDb = nλfb

D ≈ f, when a lens is used.
First minimum at 30°: b = 2λ.
Separation between mth and nth minima: (n − m)λDb.
Number of minima on one side: n ≤ bλ.
At the first minimum, wavelets from the edge and the midpoint differ by λ2, phase π.

CENTRAL MAXIMUM

7.7.4 Central maximum

Central maximum lies between the first minima on both sides

angular width = 2θ = 2λb

linear width = 2λDb = 2λfb

Half angular width = λb.
Angular width depends on λ and b, and not on D.
Decreases when slit width increases.
Increases with wavelength.

WIDTH

7.7.5 Central width equal to slit width

Central width	Then D
= b	b²2λ
= 2b	b²λ

Equal angular width changed by a percentage: λ changes by the same percentage.

Practise PYQs on this topicDiffraction at a Single Slit · 87 questions with solutions

CONCEPT

Single Slit: Secondary Maxima & Changes

In one line
Secondary maxima lie between the minima and get faint quickly. Doubling the slit width halves the angular width and makes the centre four times as bright.

SECONDARY MAXIMA

7.8.1 Secondary maxima

b sin θ = (2n + 1)λ2, n = 1, 2, 3 …

x_n = (2n + 1)λD2b

First: b sin θ = 3λ2. Second: b sin θ = 5λ2.
First minimum at sin θ = λb: first secondary maximum at sin θ = 32 × that.
Two wavelengths, same maximum: separation = (2n + 1)D Δλ2b.

COINCIDENCE

7.8.2 Two wavelengths coinciding

(2n₁ + 1)λ₁ = (2n₂ + 1)λ₂

Example: 2nd maximum of 6384 Å on 3rd maximum of λ₀: 5 × 6384 = 7λ₀.
nᵗʰ minimum of λ₁ on mᵗʰ maximum of λ₂: nλ₁ = (2m + 1)λ₂2.

INTENSITY

7.8.3 Intensity pattern

Secondary maxima: about 4.5% and 1.6% of the central maximum

First and second secondary maxima: about I₀22 and I₀61.
Fringes have unequal width and unequal intensity.
Central maximum is twice as wide as any other band.

CHANGING THE SLIT

7.8.4 Changing the slit width

Wider slit: taller and narrower central maximum

Slit width doubled	Effect
angular width	halves
amplitude at centre	doubles
intensity at centre	4 times

Half the slit covered, new λ′: width × 2 × λ′λ.
If b ≫ λ: secondary maxima disappear.

CHANGING THE COLOUR

7.8.5 Changing the colour

λ falls, so the bands become narrow and crowded.
White light: centre white, then coloured bands.

COMPARED

7.8.6 Interference and diffraction compared

	Interference	Diffraction
source	two wavefronts	one wavefront
fringe width	equal	unequal
intensity of maxima	equal	falls away
minima	may be zero	never fully dark

ENVELOPE

7.8.7 Double slit fringes inside the central maximum

number of fringes = 2λDbλDd = 2db

d = slit separation, b = width of each slit.
For n fringes inside: b = 2dn.

Practise PYQs on this topicDiffraction at a Single Slit · 87 questions with solutions

CONCEPT

Resolving Power

In one line
Diffraction blurs every point into a spot. Two points are just separated when the maximum of one spot falls on the first minimum of the other.

RAYLEIGH

7.9.1 Limit of resolution

Rayleigh’s criterion: just resolved

Limit of resolution: least separation at which two objects look distinct.
Resolving power (RP) = 1 ÷ limit of resolution.
A smaller limit means a better instrument.

TELESCOPE

7.9.2 Telescope

dθ = 1.22λD

RP = D1.22λ

D = diameter (aperture) of the objective.
RP increases with larger aperture and smaller λ.
RP does not depend on the eyepiece.
A large objective gives greater resolution.

USING dθ

7.9.3 Actual distance between two objects

x = R dθ

R = distance of the two objects from the telescope.
On the focal plane of the lens, the separation is f dθ.

MICROSCOPE

7.9.4 Microscope

d_min = 1.22λ2μ sin θ

RP = 2μ sin θ1.22λ

μ sin θ = numerical aperture (NA).
Improve RP: use a smaller λ or a higher μ (oil immersion).
RP ∝ 1λ for both instruments.
Limit of resolution ∝ λ.

HUMAN EYE

7.9.5 Human eye

x = R × 1.22λD

D = diameter of the pupil, not the radius.
R = distance of the objects from the eye.

Practise PYQs on this topicResolving Power · 27 questions with solutions

CONCEPT

Malus’s Law & Polaroids

In one line
A polaroid passes only one direction of vibration. It halves unpolarised light, and every next polaroid multiplies by cos²θ.

POLARISATION

7.10.1 Polarised and unpolarised light

Unpolarised light: E vibrates in all directions at right angles to the ray.
Plane polarised light: E vibrates in one direction only.
Only transverse waves can be polarised. Sound in air cannot.

POLAROIDS

7.10.2 Polariser and analyser

A polaroid is a thin film that passes vibrations parallel to its pass axis.
It absorbs the rest (selective absorption, dichroism).
First sheet: polariser. Second sheet: analyser.

Parallel axes pass the light; crossed axes stop it

MALUS’S LAW

7.10.3 Malus’s law

θ is the angle between the two pass axes

I = I₀ cos²θ

unpolarised light through one polaroid: I = I₀2

θ	0°	30°	45°	60°	90°
cos²θ	1	34	12	14	0

Light stopped by the first sheet is also I₀2.
Always use the angle between neighbouring sheets.
Two polaroids at 30°: I = (38)I₀ = 37.5%.

THREE POLAROIDS

7.10.4 Three polaroids

Each sheet acts on the light from the sheet before it

I = I₀2 cos²θ₁ cos²θ₂

Axes at 30° and 90° from P₁: neighbouring angles are 30° and 60°.
Each step of 60°: multiply by 14.

CROSSED

7.10.5 A sheet between crossed polaroids

I = I₀8 sin²2θ

θ = angle of the middle sheet from the first.
At 45°: maximum, I = I₀8.
Given the output: solve for sin²2θ, then θ.

n POLAROIDS

7.10.6 n polaroids at equal steps

I = I₀2 (cos²θ)^{n − 1}

n sheets at 45°: I = I₀2ⁿ.
Four sheets at 30°: three steps of 34.

Practise PYQs on this topicPolarisation · 64 questions with solutions

CONCEPT

Brewster’s Law & Other Methods

In one line
At one angle of incidence the reflected light is completely polarised, and the reflected and refracted rays are at right angles.

BREWSTER’S LAW

7.11.1 Polarisation by reflection

At the polarising angle the two rays are at 90°

tan i_p = μ

i_p + r = 90°

i_p is the polarising (Brewster) angle.
Reflected ray: completely plane polarised.
E is perpendicular to the plane of incidence.
Refracted ray: only partly polarised.
Reflected intensity is less than half of I₀.
For glass i_p ≈ 57°; for water ≈ 53°.

USING μ

7.11.2 Other forms of Brewster’s law

tan i_p = μ = cv = λ_aλ_g

λ_g = λ_a cot i_p.
i_p = cot⁻¹(vc).
μ changes with colour, so i_p is different for different colours.
Find μ from i_p, then use Snell’s law for another angle of incidence.

TWO MEDIA

7.11.3 Two media

from 1 to 2: tan i_p = μ₂μ₁

Air to glass θ_B ⇒ glass to air: 90° − θ_B.
Critical angle and Brewster angle in one question: sin C = 1μ and tan i_p = 1μ, going from denser to rarer.
From dielectric constants: μ = √ε_r.

OTHER METHODS

7.11.4 Other ways to polarise light

Dichroism: tourmaline and polaroids absorb one vibration.
Scattering: sky light at 90° to the sunlight is polarised.
Double refraction: calcite splits light into O-ray and E-ray, both polarised.

USES

7.11.5 Uses of polarisation

Sunglasses and camera filters reduce glare.
LCD displays.
3D films.
Stress analysis (photoelasticity).

ONE PLANE

7.11.6 Plane polarised waves in one plane

A² = E₁² + E₂² + 2E₁E₂ cos φ

Two plane polarised waves in the same plane add like any two waves.
E₀ and E₀ at φ = 60°: A = √3 E₀.

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Optical activity

Some substances (sugar solution, quartz) rotate the plane of polarisation.
Clockwise (looking against the light): dextro-rotatory. Anticlockwise: laevo-rotatory.

specific rotation [α] = θL × C

Polarisation by scattering

Sunlight scattered at 90° to the incident direction is plane polarised.
In other directions it is partly polarised.

Practise PYQs on this topicPolarisation · 64 questions with solutions

EXAM-DAY CARD

Just see this before the exam

I = I_max cos²φ2

intensity

β = λDd

fringe width

n = (μ − 1)tλ

film shift

2λDb

central maximum

dθ = 1.22λD

telescope

tan i_p = μ

Brewster

FORMULA SHEET

Whole chapter on one page

Wavefronts

λ′ = λμ, ν unchanged

v₁v₂ = sin isin r

Δλλ = vc

Interference

φ = 2πλ Δ

I = I₁ + I₂ + 2√I₁I₂ cos φ

I_maxI_min = (a₁ + a₂a₁ − a₂)²

YDSE

β = λDd, θ = λd

bright: nβ; dark: (n − 12)β

n₁λ₁ = n₂λ₂

Film, medium, biprism

shift = Dd(μ − 1)t

(μ − 1)t = nλ; β′ = βμ

opposite slit: λ = d²(2n − 1)D

d = √d₁d₂

Single slit

minima: b sin θ = nλ

maxima: b sin θ = (2n + 1)λ2

width = 2λDb

Resolving, polarisation

telescope RP = D1.22λ

microscope RP = 2μ sin θ1.22λ

I = I₀ cos²θ

i_p + r = 90°

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Taking nth dark at nβ

Forgetting the ½ for unpolarised light

Using 90° between non-neighbouring polaroids

Pupil radius used as D

CORRECT ✓

nth dark at (n − ½)β

First polaroid: I₀2

Use the angle between neighbours

D is the full diameter