Is it correct to say: Using \Delta = n\lambda for single-slit diffraction minima?

Single-slit *minima* occur at a sin\theta = n\lambda (n = 1, 2, ...). n = 0 is the central *maximum*.

Is it correct to say: Treating central maximum of single-slit as same width as other maxima?

Central maximum is 2\lambda D/a wide — TWICE the width of secondary maxima (\lambda D/a).

Is it correct to say: Forgetting the factor 1/2 for unpolarised light through the first polariser?

Unpolarised → polarised loses half the intensity. Malus's law only applies to *already polarised* light.

Is it correct to say: Confusing interference (from two coherent sources) with diffraction (from a single aperture)?

Interference: fringes of equal intensity and width. Diffraction: central max much brighter, side maxima fade rapidly.

Is it correct to say: Mixing up Brewster angle formula tan\theta_B = n with sin\theta_c = 1/n (critical angle)?

tan\theta_B = n is polarisation. sin\theta_c = 1/n is total internal reflection. Different phenomena.

Is it correct to say: Ignoring nm ↔ m conversion in wavelength?

1 nm = 10⁻⁹ m. Missing this factor gives answers off by 10⁹.

Wave Optics

Huygens' principle, Young's double slit, single-slit diffraction, polarization, resolving power — NCERT Class 12 Physics Ch 10

📐 13 formulas✏️ 3 examples🎯 6 practice⚖️ 5 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Young's double slit — full derivation of fringe width \beta = \lambda D/d is a 3-mark repeat every year.
→Difference between interference and diffraction (2-mark table) — memorise 4 clear points.
→Single-slit central maximum is TWICE the width of a secondary maximum. Sketching this gets 1 mark on its own.
→Brewster's law and Malus's law are separate — do not merge them.
→Always express wavelengths in metres before substituting into formulas (600 nm = 6×10⁻⁷ m).

📊 Diagram

Young's double-slit interference producing bright and dark fringes on a screen

📐 Formulas(13)

Huygens' Principle★ Board fav

Every point on a wavefront acts as a source of secondary wavelets

Path Difference and Phase Difference

Δ ϕ = \frac{2 π}{λ} Δ

Symbol	Meaning
$Δ$	Path difference (m)
$Δ ϕ$	Phase difference (rad)

Constructive Interference (bright fringe)★ Board fav

Δ = nλ, n = 0, \pm 1, \pm 2, \dots

Destructive Interference (dark fringe)

Δ = (2 n + 1) \frac{λ}{2}, n = 0, 1, 2, \dots

Resultant Intensity of Two Waves

I = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ

Fringe Width in Young's Double Slit★ Board fav

β = \frac{λ D}{d}

Symbol	Meaning
$β$	Fringe width (m)
$λ$	Wavelength (m)
$D$	Slit-to-screen distance (m)
$d$	Slit separation (m)

Position of nth Bright Fringe

y_{n} = \frac{nλ D}{d}

Position of nth Dark Fringe

y_{n} = \frac{( 2 n - 1 ) λ D}{2 d}, n = 1, 2, \dots

Single-Slit Diffraction (dark fringes)

a sin θ = nλ, n = 1, 2, 3, \dots

Width of Central Maximum (Single Slit)★ Board fav

w_{0} = \frac{2 λ D}{a}

Rayleigh Criterion (Angular Resolution)

θ_{min} = \frac{1.22 λ}{D}

Brewster's Law

tan θ_{B} = n

Malus's Law★ Board fav

I = I_{0} cos^{2} θ

✏️ Solved Examples

1Solved Exampleeasy4 steps

In a Young's double-slit experiment, light of wavelength 500 nm falls on two slits 0.4 mm apart. The screen is 1.5 m away. Find the fringe width and the position of the 3rd bright fringe from the centre.

Convert to SI units

λ = 500 \times 1 0^{- 9} m, d = 0.4 \times 1 0^{- 3} m, D = 1.5 m

2Solved Exampleboard5 steps

A parallel beam of light of wavelength 600 nm falls normally on a single slit of width 0.2 mm. A screen is 1 m away. Find (a) the width of the central bright band, (b) the linear width of the first-order secondary maximum.

Convert units

λ = 6 \times 1 0^{- 7} m, a = 2 \times 1 0^{- 4} m, D = 1 m

3Solved ExampleHOTS5 steps

Unpolarised light of intensity I_0 passes through two ideal polarisers whose axes make angle 60°. Find the transmitted intensity. If a third polariser is inserted between them at 30° to the first, find the new transmitted intensity. Compare.

First polariser transmits half of unpolarised light

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

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using \Delta = n\lambda for single-slit diffraction minima
✓Single-slit *minima* occur at a sin\theta = n\lambda (n = 1, 2, ...). n = 0 is the central *maximum*.
2
Treating central maximum of single-slit as same width as other maxima
✓Central maximum is 2\lambda D/a wide — TWICE the width of secondary maxima (\lambda D/a).
3
Forgetting the factor 1/2 for unpolarised light through the first polariser
✓Unpolarised → polarised loses half the intensity. Malus's law only applies to *already polarised* light.
4
Confusing interference (from two coherent sources) with diffraction (from a single aperture)
✓Interference: fringes of equal intensity and width. Diffraction: central max much brighter, side maxima fade rapidly.
5
Mixing up Brewster angle formula tan\theta_B = n with sin\theta_c = 1/n (critical angle)
✓tan\theta_B = n is polarisation. sin\theta_c = 1/n is total internal reflection. Different phenomena.
6
Ignoring nm ↔ m conversion in wavelength
✓1 nm = 10⁻⁹ m. Missing this factor gives answers off by 10⁹.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Wavelength = 600 nm, slit separation = 1 mm, screen distance = 2 m. Fringe width?
Q2
Central maximum of a single slit is 5 mm wide when \lambda = 500 nm, D = 1 m. Slit width?
Q3
Two polarisers are crossed (\theta = 90°). Intensity through the pair from unpolarised source of intensity I_0?
Q4
Brewster's angle for glass of n = 1.5?
Q5
In YDSE, when white light is used, the central fringe is white and other fringes are coloured. Why?
Q6
A circular aperture 3 mm across is used to view light of \lambda = 600 nm. Minimum resolvable angle?

📝 Notes

Wave Optics — key concepts

Light behaves as a wave: interference, diffraction and polarisation prove it. This chapter builds on Huygens' principle and quantifies the classic optical experiments.

Huygens' principle in one line

Every point on a wavefront acts as a source of secondary spherical wavelets; the envelope of these wavelets gives the new wavefront after time t. This single postulate reproduces the laws of reflection, refraction and diffraction.

Coherent sources

Two sources are coherent if they emit waves of the same frequency and maintain a constant phase relation. Only coherent sources produce stable interference. In practice, Young's double slit generates two coherent sources by splitting one beam.

Interference vs diffraction — the four differences

Origin: interference is from two (or more) sources; diffraction from a single aperture / obstacle.
Fringe intensity: interference — all bright fringes equally bright. Diffraction — central max is far brighter than side maxima.
Fringe width: interference — uniform. Diffraction — central max is twice the width of others.
Applicability: interference requires coherent sources; diffraction happens whenever a wavefront is truncated.

Polarisation

Only transverse waves can be polarised → confirms light is transverse.
Unpolarised light through an ideal polariser: I → I/2 (Malus's law needs already polarised input).
At Brewster's angle, reflected light is 100% polarised perpendicular to the plane of incidence.

Quick sanity checks

Interference fringe width \beta = \lambda D/d — increase D or use longer \lambda, fringes widen.
Doubling slit width in single-slit diffraction halves the central-maximum width — mnemonic: "wider slit, narrower fringe".
Resolving power of a telescope is limited by \theta_{min} = 1.22\lambda/D — larger aperture, better resolution.
Intensity through two crossed polarisers (\theta = 90°) is exactly zero; inserting a third polariser between them can increase transmission (the "quantum eraser" classic).