Wave Optics — Concepts, Formulas & Solved Numericals

Wave Optics: Why Light Behaves Like Water

When Newton said light was a stream of particles, he wasn’t entirely wrong — but he missed something beautiful. Light also travels as a wave, and this wave nature explains phenomena that particle theory simply cannot: why two overlapping beams of light can cancel each other out, why a CD shows rainbow colours, and why the sky has that bright ring around a street lamp on a foggy night.

Wave optics (also called physical optics) is the chapter that explains all of this. For Class 12 boards, it carries roughly 5 marks directly. For JEE Main, expect 1-2 questions every single attempt — and the topic appeared in JEE Main 2024 Session 1, February Session, and JEE Advanced 2023. NEET students: interference and diffraction appear almost every year, usually as single-concept application questions.

The central idea: light waves from two coherent sources can add up (constructive interference → bright fringe) or cancel out (destructive interference → dark fringe). The mathematics is elegant, and once you understand the geometry, the formulas follow naturally.

We’ll cover Huygens’ principle, interference, Young’s Double Slit Experiment (YDSE), single-slit diffraction, and polarisation — everything you need to score full marks.

Key Terms & Definitions

Wavefront — A surface joining all points that are in the same phase of oscillation. A point source gives spherical wavefronts; a source at infinity gives plane wavefronts.

Coherent sources — Two sources that maintain a constant phase difference over time. This is the prerequisite for sustained interference. Two separate bulbs are NOT coherent — their phase difference keeps changing randomly.

Path difference (Δ) — The difference in distances travelled by two waves reaching the same point. This is the single most important quantity in this chapter. Everything — bright fringes, dark fringes, maxima in diffraction — comes down to path difference.

Phase difference (δ) — Related to path difference by:

\delta = \frac{2\pi}{\lambda} \times \Delta

Fringe width (β) — The distance between two consecutive bright (or dark) fringes in YDSE. Uniform throughout the pattern.

Diffraction — The bending of light around obstacles or through slits. More pronounced when the slit width is comparable to the wavelength of light.

Polarisation — The restriction of light wave oscillations to a single plane. Transverse waves (like light) can be polarised; longitudinal waves (like sound) cannot. This fact alone is evidence that light is a transverse wave.

Core Concepts & Methods

Huygens’ Principle

Every point on a wavefront acts as a secondary source of spherical wavelets. The new wavefront is the common tangent (envelope) to all these secondary wavelets.

This principle explains reflection and refraction geometrically without invoking particles at all. For exams: Huygens’ principle is used to derive Snell’s law and to explain why wavefronts refract.

Huygens’ principle is conceptual, not numerical. CBSE loves asking 2-mark “explain using Huygens’ principle” questions. Learn to draw the wavefront diagram for refraction — it appears directly in board papers.

Young’s Double Slit Experiment (YDSE)

This is the heavyweight of this chapter. Set up: two narrow slits S₁ and S₂, separated by distance d, placed at distance D from a screen.

The geometry: For a point P at height y from the central axis, the path difference is:

\Delta = \frac{yd}{D}

This approximation holds when D >> d (which is always the case experimentally).

Condition for bright fringe (constructive interference):

\Delta = n\lambda \quad (n = 0, \pm1, \pm2, \ldots)

So bright fringes occur at: $y_n = \frac{n\lambda D}{d}$

Condition for dark fringe (destructive interference):

\Delta = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, \pm1, \pm2, \ldots)

Fringe width:

\beta = \frac{\lambda D}{d}

Position of n-th bright fringe:

y_n = \frac{n\lambda D}{d}

Position of n-th dark fringe:

y_n = \frac{(2n-1)\lambda D}{2d}

Intensity at any point (equal amplitude sources):

I = 4I_0 \cos^2\!\left(\frac{\delta}{2}\right)

where $\delta = \frac{2\pi \Delta}{\lambda}$

When a slab is introduced: If a glass slab of thickness t and refractive index μ is placed in front of one slit, the optical path increases by (μ−1)t. The central bright fringe shifts toward the slab side.

Shift in central fringe position:

$\Delta y = \frac{(\mu -1)t \cdot D}{d}$

Number of fringes shifted:

$n = \frac{(\mu -1)t}{\lambda}$

Intensity in YDSE — The Phasor Approach

For two sources with amplitudes a₁ and a₂ and phase difference δ:

I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta

When I₁ = I₂ = I₀:

I_{max} = 4I_0 \quad \text{(when } \delta = 0, 2\pi, \ldots\text{)}

I_{min} = 0 \quad \text{(when } \delta = \pi, 3\pi, \ldots\text{)}

JEE Main frequently asks: “If the intensity ratio of two sources is 4:1, find the ratio I_max : I_min.” Use the formula $\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2$ . With ratio 4:1, this gives 9:1. This exact question appeared in JEE Main 2023.

Single-Slit Diffraction

A single slit of width a gives a diffraction pattern. The central maximum is the widest and brightest. Secondary maxima are much dimmer.

Minima (dark fringes) occur at:

a\sin\theta = n\lambda \quad (n = \pm1, \pm2, \ldots)

For small angles: $\sin\theta \approx \tan\theta = y/D$

Position of n-th minimum: $y_n = \frac{n\lambda D}{a}$

Width of central maximum:

2y_1 = \frac{2\lambda D}{a}

The central maximum is twice as wide as secondary maxima.

The key distinction: in YDSE, d is the slit separation (centre to centre). In single-slit diffraction, a is the slit width. Don’t mix them up — many students use the wrong variable and get the wrong condition (bright vs dark).

Polarisation

Malus’ Law: When polarised light of intensity I₀ passes through a polariser (analyser) at angle θ to the plane of polarisation:

I = I_0 \cos^2\theta

Brewster’s Law: When unpolarised light hits a surface at Brewster’s angle (i_B), the reflected ray is completely polarised. The refracted ray is partially polarised.

\tan i_B = \mu

At Brewster’s angle, the reflected and refracted rays are perpendicular to each other.

Malus’ Law: $I = I_0 \cos^2\theta$

Brewster’s angle: $\tan i_B = \mu$

Unpolarised light through one polariser: intensity becomes $I_0/2$

Two polarisers at 90° (crossed): no light passes through

Solved Examples

Example 1 — CBSE Level

Q. In YDSE, the slits are 1 mm apart, screen is 1 m away, and wavelength is 600 nm. Find (a) fringe width, and (b) position of 3rd bright fringe.

Solution:

Given: d = 1 mm = 10⁻³ m, D = 1 m, λ = 600 nm = 600 × 10⁻⁹ m

(a) $\beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1}{10^{-3}} = 0.6 \times 10^{-3}$ m = 0.6 mm

(b) $y_3 = 3\beta = 3 \times 0.6 =$ 1.8 mm

Example 2 — JEE Main Level

Q. In YDSE with λ = 500 nm, D = 1 m, d = 0.5 mm, a glass slab of thickness 1.5 × 10⁻⁶ m and μ = 1.5 is placed over one slit. How many fringes shift? In which direction?

Solution:

Extra path introduced = (μ − 1)t = (1.5 − 1)(1.5 × 10⁻⁶) = 0.5 × 1.5 × 10⁻⁶ = 7.5 × 10⁻⁷ m

Number of fringes shifted:

$n = \frac{(\mu -1)t}{\lambda} = \frac{7.5 \times 10^{-7}}{5 \times 10^{-7}} = \mathbf{1.5 \text{ fringes}}$

Direction: The central fringe shifts toward the slab side (the slab increases optical path, so the equi-path point moves toward that slit).

The answer is 1.5 fringes — meaning the central bright fringe now sits where the 1st dark fringe used to be.

Example 3 — JEE Advanced Level

Q. Two coherent sources have intensity ratio 9:1. Find the ratio of maximum to minimum intensity in the interference pattern. Also find the ratio of slit widths.

Solution:

$I_1 : I_2 = 9 : 1$ , so $\sqrt{I_1} : \sqrt{I_2} = 3 : 1$

\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2 = \left(\frac{3+1}{3-1}\right)^2 = \left(\frac{4}{2}\right)^2 = \mathbf{4}

So $I_{max} : I_{min} = 4 : 1$ .

Intensity is proportional to slit width (since intensity ∝ amplitude² ∝ slit width for equal illumination), so slit width ratio = 9:1.

Example 4 — Polarisation (JEE Main)

Q. Unpolarised light of intensity 32 W/m² passes through three polarisers. The angle between 1st and 2nd is 30°, between 2nd and 3rd is 60°. Find final intensity.

Solution:

After 1st polariser (unpolarised → polarised): $I_1 = \frac{32}{2} = 16$ W/m²

After 2nd polariser (Malus’ law, θ = 30°): $I_2 = 16\cos^2 30° = 16 \times \frac{3}{4} = 12$ W/m²

After 3rd polariser (θ = 60° from 2nd): $I_3 = 12\cos^2 60° = 12 \times \frac{1}{4} =$ 3 W/m²

Exam-Specific Tips

CBSE Class 12

The board paper typically has:

1 mark: State Huygens’ principle / define coherent sources
2 marks: Derive expression for fringe width OR explain diffraction pattern
3 marks: YDSE numerical (find fringe width, fringe position, or shift due to slab)

For derivations, CBSE wants specific steps. For YDSE fringe width derivation, draw the geometry clearly, establish the path difference expression, then apply the condition. Marks are allocated step-wise.

CBSE 2024 asked: “Draw a ray diagram to show the propagation of a plane wavefront through a convex lens.” This is Huygens’ construction applied to refraction. Practice drawing wavefronts — not just rays.

JEE Main

Weightage: 1-2 questions per attempt. High-frequency topics:

YDSE with slab (shift calculation)
Intensity ratio problems
Malus’ law with multiple polarisers
Condition for maxima/minima when path difference involves fractions of λ

The JEE Main question pattern involves plugging into formulas with one twist — like finding the wavelength of light given fringe data, or finding how many fringes are visible.

JEE Advanced

Expect multi-concept problems: YDSE where slit widths are different (so amplitudes differ), or diffraction + interference combined (finite-width slits in YDSE). Also: qualitative questions about what changes when the entire apparatus is immersed in water (λ decreases → β decreases).

Common Mistakes to Avoid

Mistake 1: Using the wrong condition for dark fringes. Many students write $\Delta = n\lambda$ for dark fringes. Wrong. Bright fringes require $\Delta = n\lambda$ . Dark fringes require $\Delta = (n + \frac{1}{2})\lambda$ . Mixing these up costs direct marks.

Mistake 2: Forgetting the factor of 2 in central diffraction maximum width. The central maximum spans from −y₁ to +y₁, so its total width is $\frac{2\lambda D}{a}$ , not $\frac{\lambda D}{a}$ . CBSE specifically asks for “width of central maximum.”

Mistake 3: Applying Malus’ law to unpolarised light. Malus’ law ( $I = I_0\cos^2\theta$ ) only applies when the incident light is already polarised. For unpolarised light hitting a polariser, the transmitted intensity is always $I_0/2$ , regardless of orientation.

Mistake 4: Confusing fringe shift direction. When a slab is placed over S₁, the optical path of S₁ increases. The central fringe (where path difference = 0) now shifts toward S₁ — i.e., toward the slab. Students often shift it the wrong way.

Mistake 5: Using λ in the medium instead of vacuum in YDSE formulas. All YDSE formulas use the wavelength of light in the medium (λ_medium = λ_vacuum / μ). If YDSE is in water, β decreases by factor μ. If the question gives you λ in air and says the apparatus is in water — divide λ by μ before applying formulas.

Practice Questions

Q1. In YDSE, d = 0.2 mm, D = 2 m, λ = 400 nm. Find (a) fringe width and (b) number of fringes in a 1 cm wide central portion of the screen.

β = λD/d = (400 × 10⁻⁹ × 2)/(0.2 × 10⁻³) = 4 × 10⁻³ m = 4 mm

Number of fringes in 1 cm = 10 mm / 4 mm = 2.5 fringes → 2 complete fringes (count visible bright fringes: central + 1 on each side = 3, but complete fringe widths = 2)

Q2. Two slits are illuminated with light of wavelength 546 nm. The slits are 0.1 mm apart. The screen is 20 cm away. What is the angular position of the 2nd bright fringe?

For 2nd bright fringe: $d\sin\theta = 2\lambda$

$\sin\theta = \frac{2\lambda}{d} = \frac{2 \times 546 \times 10^{-9}}{0.1 \times 10^{-3}} = 0.01092$

$\theta = \sin^{-1}(0.01092) \approx 0.626°$ ≈ 0.63°

Q3. In YDSE, the intensity at the central maximum is I₀. Find the intensity at a point where the path difference is λ/6.

Phase difference δ = (2π/λ) × (λ/6) = π/3

$I = I_0 \cos^2(\delta/2) = I_0 \cos^2(\pi/6) = I_0 \times \frac{3}{4} = \frac{3I_0}{4}$

Q4. The intensity ratio of two coherent sources used in YDSE is 1:4. Find I_max : I_min.

$\sqrt{I_1} : \sqrt{I_2} = 1 : 2$

$I_{max} : I_{min} = (1+2)^2 : (2-1)^2 = 9 : 1$

Q5. In single-slit diffraction with λ = 650 nm, slit width = 0.1 mm, screen at 1 m. Find width of central maximum.

$2y_1 = \frac{2\lambda D}{a} = \frac{2 \times 650 \times 10^{-9} \times 1}{0.1 \times 10^{-3}} = 13 \times 10^{-3}$ m = 13 mm

Q6. A glass slab (μ = 1.6, t = 1.8 μm) is placed over one slit in YDSE (λ = 600 nm). By how many fringes does the pattern shift?

n = (μ−1)t/λ = (0.6 × 1.8 × 10⁻⁶)/(600 × 10⁻⁹) = (1.08 × 10⁻⁶)/(6 × 10⁻⁷) = 1.8 fringes

Q7. Polarised light of intensity I₀ passes through two polarisers. The first is at 45° to the plane of polarisation, the second is at 90° to the first. Find the final intensity.

After 1st polariser: $I_1 = I_0\cos^2 45° = I_0/2$

After 2nd polariser (90° to 1st, so 45° to I₁’s plane): Wait — the 2nd is at 90° to the 1st. So angle between I₁’s plane and 2nd polariser = 90°.

$I_2 = (I_0/2)\cos^2 90° = 0$

Final intensity = 0 (crossed polarisers)

Q8. At what angle should a ray of light strike a glass surface (μ = 1.73) so that the reflected ray is completely plane-polarised?

Using Brewster’s law: $\tan i_B = \mu = 1.73 = \tan 60°$

Brewster’s angle = 60°

FAQs

Why do we need coherent sources for interference? Can’t any two light sources interfere?

Any two waves can superpose, but to see a stable interference pattern, the phase difference at every point must remain constant over time. A normal bulb changes phase millions of times per second — the bright and dark fringes keep shifting so fast you see uniform average intensity. Coherent sources (same frequency, constant phase difference) give a pattern that stays put and is visible.

What is the difference between interference and diffraction?

Both involve superposition of waves. Interference is the superposition of waves from discrete sources (like two slits). Diffraction is superposition of wavelets from different parts of the same wavefront (like parts of a single slit). In practice, every real interference experiment also involves diffraction — YDSE shows interference fringes modulated by the single-slit diffraction envelope.

Why does fringe width decrease when the slit separation increases?

$\beta = \lambda D/d$ . As d increases, the path difference between the two slits for the same screen position increases. So the same path difference of λ (first bright fringe) is reached at a smaller y. The fringes squeeze together.

What happens to the YDSE pattern if white light is used instead of monochromatic light?

The central fringe remains white (all wavelengths have zero path difference at centre). On either side, the fringes fan out into colour-dispersed bands — violet closest to centre (smallest λ, smallest β), red farthest. After a few orders, the fringes of different colours overlap and you see white again.

Why can sound not be polarised but light can?

Polarisation requires the wave oscillation to be in a transverse direction (perpendicular to propagation). Light is a transverse electromagnetic wave — its electric field can be restricted to one plane. Sound is a longitudinal wave — particles oscillate along the direction of propagation. There is no “perpendicular plane” to restrict, so polarisation has no meaning for sound.

What is the significance of Brewster’s angle in practical applications?

Camera filters use polarising filters to remove glare from water and glass surfaces — light reflected near Brewster’s angle is almost fully polarised, and a polariser at 90° to it blocks the glare. LCD screens use crossed polarisers: the liquid crystal layer rotates the plane of polarisation selectively, turning pixels on and off.

If the entire YDSE apparatus is immersed in water, what changes?

The wavelength of light in water is λ_water = λ_air/μ_water ≈ λ/1.33. Since β = λD/d, the fringe width decreases by a factor of 1.33. The pattern becomes more compact. The number of visible fringes may increase if the screen is large enough.

Why is the central maximum in single-slit diffraction twice as wide as the secondary maxima?

The first minimum occurs when the path difference across the full slit width equals λ (i.e., asinθ = λ). But the central maximum spans both sides of zero — from −θ₁ to +θ₁. Each secondary maximum only spans from one minimum to the next (one half-period of the pattern), which is half the width of the central maximum.