Wave optics for JEE Main

Wave optics questions in JEE Main are compact. Most are Young's double-slit experiment in some variation: change the wavelength, the medium, the slit separation or the intensity of one slit, or cover a slit with a thin film. A smaller share comes from single-slit diffraction and polarisation. Learn a handful of results thoroughly and know exactly what each depends on.

The official 2026 syllabus covers wavefronts and Huygens' principle, reflection and refraction explained with Huygens' principle, interference, Young's double-slit experiment and the fringe width, coherent sources and sustained interference, diffraction at a single slit and the width of the central maximum, and polarisation: plane-polarised light, Brewster's law, and uses of plane-polarised light and Polaroids.

Wavefronts and Huygens' principle

A wavefront is a surface of constant phase: spherical near a point source, plane far away. Huygens' principle says every point on a wavefront acts as a source of secondary wavelets, and the new wavefront is the common tangent to these wavelets a moment later.

Applied to a plane wave hitting a surface, it reproduces the laws of reflection and Snell's law, with n = c/v. In a denser medium light slows down and its wavelength shortens to λ/n, while its frequency does not change.

Interference and coherent sources

When two waves overlap, their displacements add. For a steady (sustained) pattern the sources must be coherent: same frequency and a constant phase difference. Two separate bulbs are never coherent, so practical set-ups split light from one source, as the two slits of Young's experiment do.

With intensities I₁ and I₂ and phase difference φ:

I = I₁ + I₂ + 2√(I₁I₂) cos φ

For equal intensities I₀, this becomes I = 4I₀ cos²(φ/2). Energy is not lost; it is redistributed from the dark fringes to the bright ones.

I_max/I_min = (√I₁ + √I₂)² / (√I₁ − √I₂)²

Young's double-slit experiment

Slits a distance d apart, a screen at distance D (D much larger than d). At a point y from the centre, the path difference is yd/D.

A film adds an optical path (n − 1)t to one beam. The central bright fringe moves to where the geometric path difference cancels it, which is on the side of the covered slit. The fringe width is unchanged.

With white light, the central fringe is white (every colour has zero path difference there) and the fringes on either side are coloured.

Diffraction at a single slit

Light passing through a single slit of width a spreads out. Minima occur where

a sin θ = nλ (n = 1, 2, 3 …)

The central maximum stretches between the first minima on either side, so its angular width is 2λ/a and its width on a screen at distance D is 2λD/a. The secondary maxima lie roughly midway between minima and are much fainter.

A narrower slit or a longer wavelength gives a wider central maximum.

Polarisation

Only transverse waves can be polarised, so polarisation shows that light is transverse.

• A Polaroid passes only the component of vibration along its transmission axis. Unpolarised light of intensity I₀ comes out with I₀/2, whatever the orientation.
• A second Polaroid at angle θ to the first passes I cos²θ of the polarised light reaching it (Malus's law).
• Brewster's law: light reflected at the angle θ_B where tan θ_B = n is completely plane polarised, vibrating perpendicular to the plane of incidence. At this angle the reflected and refracted rays are at 90°, so θ_B + r = 90°.

Uses include polarising sunglasses that cut glare from water and roads, camera filters, 3D film glasses and liquid-crystal displays.

Worked problems

Problem 1: fringe positions. In a YDSE, d = 0.5 mm, D = 1 m and λ = 600 nm. Find the fringe width and the distance of the third dark fringe from the centre.

β = (600 × 10⁻⁹ × 1)/(0.5 × 10⁻³) = 1.2 × 10⁻³ m = 1.2 mm.
Third dark fringe: path difference 5λ/2, so y = 2.5β = 3 mm.

Problem 2: unequal slits. The intensities from the two slits are in the ratio 9 : 1. Find I_max/I_min.

Amplitudes are in the ratio 3 : 1, so I_max/I_min = (3 + 1)²/(3 − 1)² = 16/4 = 4.

Problem 3: thin film. A film of refractive index 1.5 over one slit shifts the pattern by 5 fringes for λ = 600 nm. Find its thickness.

(n − 1)t = 5λ, so t = (5 × 600 × 10⁻⁹)/0.5 = 6 × 10⁻⁶ m = 6 μm.

Problem 4 (numerical answer): central maximum. Light of wavelength 500 nm falls on a slit 0.1 mm wide. Find the width of the central maximum, in mm, on a screen 2 m away.

2λD/a = (2 × 500 × 10⁻⁹ × 2)/(10⁻⁴) = 0.02 m = 20 mm.

Practice set

1. A YDSE set-up is moved from air into water (n = 4/3). What happens to the fringe width?
2. What fraction of unpolarised light passes through a single Polaroid?
3. Find the Brewster angle for glass of n = √3, and the angle of refraction at that incidence.
4. Unpolarised light of intensity I₀ passes through two Polaroids whose axes are at 60°. Find the final intensity.
5. In a YDSE, the slit separation is doubled and the screen distance halved. What happens to the fringe width?
6. Two coherent sources, each of intensity I₀, meet with a phase difference of π/2. Find the resultant intensity.
7. Light of wavelength 600 nm in air enters glass of n = 1.5. Find its wavelength, frequency and speed in glass (c = 3 × 10⁸ m/s).
8. Ten fringes fit in a certain width on the screen for λ = 600 nm. How many fit in the same width for λ = 400 nm?

Answers

1. It becomes 3/4 of its value in air.
2. One half.
3. tan θ_B = √3, so θ_B = 60°; r = 90° − 60° = 30°.
4. I₀/2 × cos²60° = I₀/2 × 1/4 = I₀/8.
5. β = λD/d becomes one quarter.
6. 4I₀ cos²(π/4) = 2I₀.
7. λ = 600/1.5 = 400 nm; frequency unchanged at 3 × 10⁸/(6 × 10⁻⁷) = 5 × 10¹⁴ Hz; speed 2 × 10⁸ m/s.
8. β ∝ λ, so fringes are narrower by 2/3 and 15 fit.

What to do next

• Make one table of what changes β in a YDSE: λ, D, d, the medium, and a film over one slit.
• Solve 15 YDSE problems, including intensity at a given point and the thin-film shift.
• Practise single-slit and Polaroid questions until the factor of two in each is automatic.
• Revise ray optics, then move on to modern physics.