Wave Optics NEET PYQs (2005–2026) with Solutions

10 solved Wave Optics PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumInterference

   In Young’s double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ is K units. The intensity of light at a point where the path difference is λ/3 will be:

   1. AOption A: K/4
   2. BOption B: K
   3. COption C: 2K
   4. DOption D: K/2

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   Correct answer: (A) K/4

   Explanation

   Intensity in YDSE is given by: I = I₀ cos²(φ/2) where phase difference: φ = (2π/λ)Δx For path difference Δx = λ: φ = 2π Therefore: I = I₀ cos²(π) I = I₀ Given: K = I₀ Now for path difference: Δx = λ/3 Phase difference: φ = (2π/λ)(λ/3) φ = 2π/3 Hence: I = I₀ cos²(π/3) I = I₀(1/2)² I = I₀/4 Since K = I₀: I = K/4 Hence option (A) is correct.

2. Question 2 (NEET 2026)

   NEET 2026EasyInterference

   In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy. B. Diffraction and interference are characteristics exhibited only by light waves. Choose the correct answer from the options given below:

   1. AOption A: A is true and B is also true
   2. BOption B: A is false, but B is true
   3. COption C: A is true, but B is false
   4. DOption D: Both A and B are false

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   Correct answer: (C) A is true, but B is false

   Explanation

   In interference and diffraction, energy is redistributed and conserved, so statement A is true. Interference and diffraction occur in all wave phenomena including sound waves, hence statement B is false. Therefore, option (C) is correct.

3. Question 3 (NEET 2025)

   NEET 2025MediumPolarisation

   An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster’s angle. Then-

   1. AOption A: reflected light is completely polarized and the angle of reflection is close to 60°
   2. BOption B: reflected light is partially polarized and the angle of reflection is close to 30°
   3. COption C: both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to 60° and 30°, respectively
   4. DOption D: transmitted light is completely polarized with angle of refraction close to 30°

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   Correct answer: (A) reflected light is completely polarized and the angle of reflection is close to 60°

   Explanation

   At Brewster’s angle, reflected light becomes completely plane polarized. Using Brewster’s law: $$\mu = \tan i_B$$ $$1.73 \approx \tan i_B$$ Hence, $$i_B \approx 60^\circ$$. Since angle of reflection equals angle of incidence, the reflected angle is also close to 60°.

4. Question 4 (NEET 2025)

   NEET 2025MediumPolarisation

   The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at 22.5° from the polarization axis of one of the polaroid, is ($I_0$ is the intensity of polarized light after passing through the first polaroid):

   1. AOption A: $\dfrac{I_0}{2}$
   2. BOption B: $\dfrac{I_0}{4}$
   3. COption C: $\dfrac{I_0}{8}$
   4. DOption D: $\dfrac{I_0}{16}$

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   Correct answer: (C) $\dfrac{I_0}{8}$

   Explanation

   Using Malus’ law twice: First transmission through the middle polaroid: $$I_1 = I_0 \cos^2 22.5^\circ$$ The angle between the middle polaroid and the final crossed polaroid is also $67.5^\circ$.$$I = I_1 \cos^2 67.5^\circ$$Since:$$\cos^2 22.5^\circ \times \cos^2 67.5^\circ = \frac{1}{8}$$Hence,$$I = \frac{I_0}{8}$$ Therefore, option (C) is correct.

5. Question 5 (NEET 2024)

   NEET 2024EasyInterference

   If the monochromatic source in Young’s double slit experiment is replaced by white light, then

   1. AOption A: there will be a central bright white fringe surrounded by a few coloured fringes.
   2. BOption B: all bright fringes will be of equal width.
   3. COption C: interference pattern will disappear.
   4. DOption D: there will be a central dark fringe surrounded by a few coloured fringes.

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   Correct answer: (A) there will be a central bright white fringe surrounded by a few coloured fringes.

   Explanation

   In Young's double slit experiment using white light, all wavelengths constructively interfere at the center, producing a bright white central fringe. Away from the center, different wavelengths produce coloured fringes. Hence, option (A) is correct.

6. Question 6 (NEET 2024)

   NEET 2024MediumPolarisation

   An unpolarised light beam strikes a glass surface at Brewster’s angle. Then:

   1. AOption A: both the reflected and refracted light will be completely polarised.
   2. BOption B: the reflected light will be completely polarised but the refracted light will be partially polarised.
   3. COption C: the reflected light will be partially polarised.
   4. DOption D: the refracted light will be completely polarised.

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   Correct answer: (B) the reflected light will be completely polarised but the refracted light will be partially polarised.

   Explanation

   At Brewster’s angle, the reflected light is completely plane polarised. The refracted light is only partially polarised. Hence, option (B) is correct.

7. Question 7 (NEET 2023)

   NEET 2023MediumInterference

   For Young’s double slit experiment, two statements are given below: Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant. Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases. In the light of the above statements, choose the correct answer from the options given below:

   1. AOption A: Both Statement I and Statement II are true.
   2. BOption B: Both Statement I and Statement II are false.
   3. COption C: Statement I is true but Statement II is false.
   4. DOption D: Statement I is false but Statement II is true.

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   Correct answer: (C) Statement I is true but Statement II is false.

   Explanation

   Angular fringe separation: θ = λ/d It does not depend on screen distance, so Statement I is true. If wavelength increases, angular separation also increases, not decreases. Hence Statement II is false. Therefore, option (C) is correct.

8. Question 8 (NEET 2022)

   NEET 2022MediumInterference

   In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is:

   1. AOption A: 12
   2. BOption B: 6
   3. COption C: 8
   4. DOption D: 9

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   Correct answer: (A) 12

   Explanation

   Fringe width in YDSE: β ∝ λ Number of fringes in a fixed screen length: N ∝ 1/λ Thus, N₂/N₁ = λ₁/λ₂ N₂/8 = 600/400 N₂ = 12 Hence option (A) is correct.

9. Question 9 (NEET 2020)

   NEET 2020MediumPolarisation

   The Brewster angle i$_b$ for an interface should be

   1. AOption A: 0° < i$_b$ < 30°
   2. BOption B: 30° < i$_b$ < 45°
   3. COption C: 45° < i$_b$ < 90°
   4. DOption D: i$_b$ = 90°

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   Correct answer: (C) 45° < i$_b$ < 90°

   Explanation

   Brewster angle satisfies tan i₍b₎ = μ. Since refractive index μ > 1 for ordinary media, Brewster angle is always greater than 45° and less than 90°.

10. Question 10 (NEET 2019)

    NEET 2019MediumDiffraction

    Angular width of the central maxima in the Fraunhofer diffraction for λ = 6000 Å is θ₀. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is:

    1. AOption A: 420 Å
    2. BOption B: 1800 Å
    3. COption C: 4200 Å
    4. DOption D: 6000 Å

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    Correct answer: (C) 4200 Å

    Explanation

    For Fraunhofer diffraction, angular width of central maxima: θ = 2λ/a Initially: θ₀ = 2 × 6000/a New angular width decreases by 30%: θ' = 0.7θ₀ Since θ ∝ λ, λ'/6000 = 0.7 λ' = 4200 Å