503 questions on Optics for NEET 2027, 2 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.
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Answer: (A) Wavelength and slit separation
β = λD/d. Fringe width ∝ λ (wavelength) and D (screen distance), inversely ∝ d (slit separation).
Quick tip: β = λD/d. Increase λ or D → wider fringes. Increase d → narrower fringes.
NCERT Class 12, Chapter 10, page 362
Answer: (B) 20°
When a plane mirror is rotated by an angle θ about an axis in the plane of the mirror and perpendicular to the plane of incidence, keeping the incident ray fixed, the reflected ray turns through an angle of 2θ. Here θ = 10°, so the reflected ray is deflected by 2 × 10° = 20°.
Answer: (B) 26.7 cm
1/v + 1/u = 1/f with u = −80 cm and f = −20 cm gives v = −26.7 cm: a real image 26.7 cm in front of the mirror.
Answer: (C) 2/√3
Since the ray enters normally through face AB, it proceeds undeviated to face AC. The geometry of triangle ABC gives angle A = 60°, so the angle made with face AC is 30°, which means the angle of incidence at face AC is r = 90° - 30° = 60°. For total internal reflection to occur, r ≥ θc, which implies sin(60°) ≥ 1/μ. Thus, μ ≥ 1/sin(60°) = 2/√3.
Answer: (B) 37.5%
The first polaroid transmits an intensity I₁ = I₀/2. According to Malus's law, the final transmitted intensity is I = I₁ cos²(30°) = (I₀/2) × (√3/2)² = (I₀/2) × (3/4) = (3/8) I₀ = 0.375 I₀, which corresponds to 37.5%.
Answer: (D) 3 cm
The apparent depth is d' = d / μ = 12 / (4/3) = 9 cm. Therefore, the apparent shift (the distance the coin appears raised) is Δd = d - d' = 12 - 9 = 3 cm.
Answer: (A) √2
For TIR at the curved surface, r' ≥ C, where r' = 90° - r. Thus, cos r ≥ sin C = 1/n, which means 1 - sin² r ≥ 1/n². Using Snell's law at the entrance face, sin r = (sin θ)/n. This gives 1 - (sin² θ)/n² ≥ 1/n², or n² ≥ 1 + sin² θ. The most demanding condition occurs when sin θ = 1 (grazing incidence), yielding n² ≥ 2, or n ≥ √2.
Answer: (B) 1.2 μm
For the first minimum in single-slit diffraction, a sin θ = 1 × λ. Given θ = 30° and λ = 600 nm = 0.6 μm, we have a sin 30° = 0.6 μm, so a × 0.5 = 0.6 μm, which gives a = 1.2 μm.
Answer: (B) At a depth of 6 cm below the curved apex (no shift)
The mark is at the center of curvature of the spherical curved surface. Any ray originating from the center of curvature hits the spherical boundary normally (angle of incidence = 0°), so it passes undeviated. Thus, the apparent position coincides with the actual position, at 6 cm below the vertex.
Answer: (D) I₀ / 22
The intensity in single-slit diffraction is given by I = I₀ [sin(β) / β]², where β = (π a sinθ) / λ. For the first secondary maximum, β ≈ 3π/2. Substituting this gives I₁ ≈ I₀ [sin(3π/2) / (3π/2)]² = I₀ [ -1 / (1.5π) ]² = I₀ / (2.25 × 9.87) ≈ I₀ / 22.2 (about 4.5% of I₀).
Answer: (D) 4 D
P = P₁ + P₂ = 100/50 + 100/(50) = 4 D (focal lengths in cm, so P = 100/f).
Answer: (C) 7.5 D
P = P₁ + P₂ = 100/40 + 100/(20) = 7.5 D (focal lengths in cm, so P = 100/f).
Answer: (D) Virtual, erect, and magnification +0.60
For a convex mirror, f = +R/2 = +30 cm. Object distance u = -20 cm. Using the magnification formula m = f / (f - u) = 30 / [30 - (-20)] = 30 / 50 = +0.60. A positive magnification indicates that the image is virtual and erect.
Answer: (B) 0.86 cm
When the final image is formed at infinity, the intermediate real image is located at the first focus of the eyepiece, so the image distance from the objective is v₀ = 14.0 - 2.0 = 12.0 cm. Using the lens formula 1/v₀ - 1/u₀ = 1/f₀, we get 1/12.0 - 1/u₀ = 1/0.8, which gives 1/u₀ = 1/12 - 1/0.8 = (1 - 15)/12 = -14/12. Thus, |u₀| = 12/14 ≈ 0.857 cm ≈ 0.86 cm.
Answer: (C) Statement I is incorrect but Statement II is correct
Statement I is incorrect because if the refractive index of the surrounding medium is greater than that of the lens, (μ_lens / μ_med - 1) becomes negative, changing a convex lens into a diverging lens. Statement II is correct as the sign of the focal length reverses when μ_med > μ_lens.