Wave Optics NEET PYQReal NEET & AIPMT 2008–2020 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Wave Optics question from the NEET & AIPMT papers on this site (2008–2020), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Wave Optics previous year questions
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Q1NEET 2020 · Q138
The Brewsters angle i_b for an interface should be :
Not quite — the answer is C.
Brewster's law gives tan i_b = μ. For light going from a rarer to a denser medium μ > 1, so tan i_b > 1 and i_b lies between 45° and 90°. i_b = 90° would need an infinite refractive index.
Q2NEET 2020 · Q165
In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :
Not quite — the answer is C.
Fringe width β = λD/d. Doubling D and halving d multiplies β by 2 × 2 = 4. The two changes reinforce each other rather than cancelling.
Q3NEET 2020 · Q168
Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is :
Not quite — the answer is A.
The angular limit of resolution is Δθ = 1.22λ/D = 1.22 × 600×10⁻⁹ / 2 = 3.66×10⁻⁷ rad. Leaving out the factor 1.22 gives 3.0×10⁻⁷, and using the radius instead of the diameter doubles the answer to 7.32×10⁻⁷.
Q4NEET 2019 · Q5
In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (μ_water = 4/3)
Not quite — the answer is B.
Angular width ∝ λ, and in water λ becomes λ/μ. New width = 0.2° × 3/4 = 0.15°. Multiplying by 4/3 (0.266°) wrongly assumes the wavelength increases in water.
Q5NEET 2018 · Q8
Unpolarised light is incident from air on a plane surface of a material of refractive index 'μ'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
Not quite — the answer is B.
This is Brewster's angle, tan i = μ, at which the reflected light is completely plane polarised with its electric vector perpendicular to the plane of incidence. The expression with tan⁻¹(1/μ) inverts Brewster's law, and sin⁻¹(1/μ) is the critical angle.
Q6NEET 2018 · Q9
In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed to
Not quite — the answer is B.
Angular fringe width θ = λ/d, so θ ∝ 1/d. d′ = d × 0.20/0.21 = 2 × 0.952 ≈ 1.9 mm. Increasing d to 2.1 mm would narrow the fringes instead.
Q7NEET 2017 (Gujarati paper) · Q154
Two coherent sources of intensity ratio α interfere. The value of (I_max − I_min)/(I_max + I_min) is:
Not quite — the answer is A.
Let I₁ = αI₂. I_max = (√I₁ + √I₂)² and I_min = (√I₁ − √I₂)², so I_max − I_min = 4√(I₁I₂) and I_max + I_min = 2(I₁ + I₂). The ratio is 2√(I₁I₂)/(I₁ + I₂) = 2√α/(1 + α). The inverse form (1 + α)/(2√α) is always at least 1, which a ratio of this kind cannot be.
Q8NEET 2017 (Gujarati paper) · Q166
Due to Doppler effect, the shift in wavelength observed is 0.1 Å, for a star producing a wavelength 6000 Å. The velocity of recession of the star will be:
Not quite — the answer is C.
For v ≪ c, Δλ/λ = v/c, so v = c × Δλ/λ = 3 × 10⁸ × 0.1/6000 = 5 × 10³ m s⁻¹ = 5 km s⁻¹. Doubling the shift by mistake would give 10 km s⁻¹.
Q9NEET 2017 (Gujarati paper) · Q173
A parallel beam of light of wavelength λ is incident normally on a single slit of width d. Diffraction bands are obtained on a screen placed at a distance D from the slit. The second dark band from the central bright band will be at a distance given by:
Not quite — the answer is D.
Single-slit minima occur at d sin θ = nλ. For n = 2 and small angles, y = D tan θ ≈ 2λD/d. The form 2λd/D has the slit width and screen distance swapped and does not even have units of length.
Q10NEET 2016 Phase 2 · Q36
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio (I_max − I_min)/(I_max + I_min) will be
Not quite — the answer is B.
With I₁ = nI₂, I_max = (√I₁ + √I₂)² and I_min = (√I₁ − √I₂)². Their difference is 4√(I₁I₂) and sum is 2(I₁ + I₂), so the ratio is 2√(I₁I₂)/(I₁ + I₂) = 2√n/(n + 1). Dropping the factor 2 gives √n/(n + 1).
Q11NEET 2016 Phase 2 · Q38
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10⁻⁵ cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is
Not quite — the answer is D.
For single-slit diffraction the first minimum is at sinθ = λ/a, and on the focal plane y = fλ/a = 60 × 5 × 10⁻⁵/0.02 = 0.15 cm. Using 2λ/a would give the width of the central maximum (0.30 cm), not the position of the first dark band.
Q12NEET 2016 Phase 1 · Q6
In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of:
Not quite — the answer is D.
First minimum: a sin30° = λ, so λ/a = 1/2. The first secondary maximum lies at a sinθ ≈ 3λ/2, giving sinθ = 3/2 × 1/2 = 3/4. sin⁻¹(1/2) is the first minimum itself, not a maximum.
Q13NEET 2016 Phase 1 · Q18
The intensity at the maximum in a Young's double slit experiment is I₀. Distance between two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d?
Not quite — the answer is D.
The point in front of a slit is at y = d/2, so path difference = yd/D = (d/2)(d)/(10d) = d/20 = λ/4, i.e. phase difference π/2. I = I₀cos²(φ/2) = I₀cos²(π/4) = I₀/2. Using I₀cos²φ instead of cos²(φ/2) would wrongly give zero.
Q14AIPMT 2015
In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?
Not quite — the answer is B.
Fringe width β = λD/d = 5 × 10⁻⁷ × 1/10⁻³ = 0.5 mm. Ten fringes fill the central diffraction maximum: 2λD/a = 10β = 5 mm, so a = 2 × 5 × 10⁻⁷ × 1/(5 × 10⁻³) = 2 × 10⁻⁴ m = 0.2 mm.
Q15AIPMT 2015
For a parallel beam of monochromatic light of wavelength 'λ', diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the width of the central maxima will be:
Not quite — the answer is B.
The first minima are at sin θ ≈ λ/a on either side, so the central maximum spans 2Dλ/a on the screen. Dλ/a is only the half-width.
Q16AIPMT 2014
A beam of light of λ = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is:
Not quite — the answer is C.
First minima lie at y = ±λD/a, so their separation is 2λD/a = 2 × 600 × 10⁻⁹ × 2/10⁻³ = 2.4 × 10⁻³ m = 2.4 mm. Forgetting the factor 2 gives 1.2 mm.
Q17AIPMT 2014
In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ/4, will be:
Not quite — the answer is B.
Path difference λ is a maximum, so K = I_max. Path difference λ/4 means phase difference π/2, so I = I_max cos²(π/4) = K/2.
Q18NEET 2013
In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ₁ = 12000 Å and λ₂ = 10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?
Not quite — the answer is B.
Fringe width β = λD/d: β₁ = (12000 × 10⁻¹⁰ × 2)/(2 × 10⁻³) = 1.2 mm and β₂ = 1.0 mm. Bright fringes coincide when n₁(1.2) = n₂(1.0), first at n₁ = 5, n₂ = 6, i.e. 6 mm from the centre.
Q19NEET 2013
A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct ?
Not quite — the answer is C.
Faster electrons have a smaller de Broglie wavelength λ = h/(mv). The angular width of the central maximum is about 2λ/a, so it decreases. Electrons do show diffraction because of their wave nature.
Q20AIPMT 2008 Prelims · Q22
Two periodic waves of intensities I₁ and I₂ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
Not quite — the answer is A.
On superposition, Imax = (√I₁ + √I₂)² and Imin = (√I₁ − √I₂)². Adding them, the cross terms ±2√(I₁I₂) cancel, leaving 2(I₁ + I₂). (√I₁ + √I₂)² is only the maximum, not the sum.
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