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JEE Main Wave Optics — practice questions

Explore 125 free Wave Optics questions in JEE Main Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Wave Optics: quick revision

Interference depends on phase difference between coherent waves. In Young’s double-slit setup, the small-angle fringe relation assumes a distant screen. Diffraction spreads light from a finite aperture and has different minimum conditions from two-slit interference.

Key ideas and when to use them

  • Double-slit fringe width: $\beta=\lambda D/d$ for $D\gg d$ and small angles.
  • Phase difference: $\delta=2\pi\Delta/\lambda$.
  • Single-slit first minimum: $a\sin\theta=\lambda$.

Common mistakes

  • Keep slit separation d distinct from single-slit width a.
  • Convert nanometres and millimetres to metres before combining values.

Three worked examples

Example 1: In a double-slit setup, λ = 500 nm, screen distance D = 2 m and slit separation d = 0.50 mm. Find the fringe width.
  1. Use $\beta=\lambda D/d$.
  2. $\beta=(500\times10^{-9})2/(0.50\times10^{-3})$.

0.002 m, or 2 mm

Example 2: Two coherent waves of wavelength 600 nm have path difference 300 nm. Find their phase difference as a multiple of π.
  1. Use $\delta=2\pi\Delta/\lambda$.
  2. $\delta=2\pi\times300/600=\pi$.

π rad

Example 3: Light of wavelength 500 nm passes through a slit of width 0.20 mm. Find the small angle of the first diffraction minimum.
  1. For the first minimum, $\sin\theta=\lambda/a$.
  2. At small angles, $\theta\approx(500\times10^{-9})/(0.20\times10^{-3})$.

0.0025 rad

Qallery AI-assisted study notes and original examples. Further reading: NCERT Physics — Wave Optics.

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Wave Optics questions and exercises with answers

Showing questions 1–30 of 125.

Unpolarised light of intensity $I_0$ is passed through a single ideal polariser. The intensity of the transmitted light is:Read the answer and worked solution →Brewster's angle for a glass of refractive index $\sqrt{3}$ (relative to air) is:Read the answer and worked solution →The SI unit of electric charge is the:Read the answer and worked solution →Coulomb's law: force between two point charges is proportional to:Read the answer and worked solution →The value of k in Coulomb's law (in SI) is approximately:Read the answer and worked solution →Electric field strength at a point is defined as force per unit:Read the answer and worked solution →Total electric flux through a closed surface enclosing charge q (Gauss's law) is:Read the answer and worked solution →Electric field due to an INFINITE SHEET of charge with surface charge density σ:Read the answer and worked solution →Two point charges +2μC and -3μC are separated by 30 cm. The force on each (k = 9 × 10⁹):Read the answer and worked solution →Electric field due to a point charge q at distance r:Read the answer and worked solution →Critical angle for total internal reflection from water (n = 4/3) to air:Read the answer and worked solution →In a thin film of thickness t with refractive index n, condition for destructive interference for reflected light (no phase change on reflection conditions):Read the answer and worked solution →Diffraction is the:Read the answer and worked solution →Single-slit diffraction first minimum occurs at angle θ where:Read the answer and worked solution →Brewster's angle θ_B for light hitting glass (n = 1.5) from air satisfies:Read the answer and worked solution →For maximum brightness in Newton's rings (reflected light, air film), what is the radius of n-th bright ring? (R = lens radius, λ = wavelength)Read the answer and worked solution →Two slits 0.5 mm apart with light of wavelength 5000 Å. The 5th bright fringe is at distance from central maximum on screen 1 m away:Read the answer and worked solution →In Young's experiment, if the entire setup is immersed in water (n = 4/3), the fringe width:Read the answer and worked solution →Power of a convex lens of focal length 50 cm:Read the answer and worked solution →Resolution limit of a telescope by Rayleigh criterion:Read the answer and worked solution →Wavelength of light in water (n = 4/3) if its vacuum wavelength is 600 nm:Read the answer and worked solution →Two coherent sources have intensities I1 = 9I and I2 = I. Maximum and minimum intensities in interference pattern:Read the answer and worked solution →In Young's experiment, intensity at a point where path difference is λ/4:Read the answer and worked solution →For a grating with 500 lines/mm and light at 600 nm, find angle for the first-order maximum:Read the answer and worked solution →In single-slit diffraction, ratio of intensities of first and central maxima is approximately:Read the answer and worked solution →Speed of light in a medium of refractive index 1.5 is:Read the answer and worked solution →A wavefront is:Read the answer and worked solution →Young's double-slit experiment demonstrates:Read the answer and worked solution →For constructive interference, path difference Δp =Read the answer and worked solution →In YDSE, fringe width β =Read the answer and worked solution →

Worked practice questions

Unpolarised light of intensity $I_0$ is passed through a single ideal polariser. The intensity of the transmitted light is:

  1. $I_0$
  2. $I_0/2$
  3. $I_0\cos^2\theta$
  4. $I_0/4$
Show answer and solution

Correct answer: B. $I_0/2$

Unpolarised light contains all polarisation directions with equal probability. Averaging Malus's law over all input angles gives $\langle I \rangle = I_0 \langle\cos^2\theta\rangle = I_0/2$. Once polarised, a second polariser would then attenuate further by $\cos^2\theta$ relative to its axis.

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Brewster's angle for a glass of refractive index $\sqrt{3}$ (relative to air) is:

  1. $30^\circ$
  2. $45^\circ$
  3. $60^\circ$
  4. $90^\circ$
Show answer and solution

Correct answer: C. $60^\circ$

**Brewster's law:** $\tan i_B = \mu$. With $\mu = \sqrt{3}$, $\tan i_B = \sqrt{3} \Rightarrow i_B = 60^\circ$. At this angle the reflected light is completely plane-polarised, with the electric field perpendicular to the plane of incidence.

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The SI unit of electric charge is the:

  1. volt
  2. ampere
  3. coulomb
  4. ohm
Show answer and solution

Correct answer: C. coulomb

Charge is measured in coulombs (C). 1 C = 1 A·s. The fundamental unit of charge is e = 1.6 × 10⁻¹⁹ C (proton/electron magnitude).

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