50 questions on this topic of Optics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (A) It remains unchanged.
The linear width of the central maximum is given by w = 2λD / a. When both D and a are doubled, w' = 2λ(2D) / (2a) = 2λD / a = w. Therefore, the linear width remains unchanged.
Answer: (D) 7 : 8
For the 4th minimum of λ₁, a sin θ = 4λ₁. For the 3rd secondary maximum of λ₂, a sin θ = (3 + 1/2)λ₂ = 3.5λ₂ = (7/2)λ₂. Equating the two gives 4λ₁ = (7/2)λ₂, which leads to λ₁ / λ₂ = 7 / 8.
Answer: (D) 6 zones; a diffraction minimum
When the path difference across the slit is 3λ, the slit can be divided into 2n = 2(3) = 6 equal zones, each with a relative edge-to-edge path difference of λ/2. Successive pairs of adjacent zones cancel each other out pairwise (1 cancels 2, 3 cancels 4, 5 cancels 6), producing net zero amplitude, which corresponds to the third diffraction minimum.
Answer: (A) Both Statement I and Statement II are correct
Statement I is correct because the angular width of the central maximum is 2λ/a, whereas each secondary maximum has an angular width of λ/a, which is half of 2λ/a. Statement II is also correct because the intensity factor (sin β)/β decreases with increasing β, shifting the positions of the maxima slightly toward smaller angles compared to the exact midpoints ((2n+1)π/2).
Answer: (D) 1.0 μm
The condition for diffraction minima is a·sinθ = nλ. For the first minimum, n = 1, so a·sin(30°) = 1 × 500 nm. Since sin(30°) = 0.5, we get a = 500 nm / 0.5 = 1000 nm = 1.0 μm.
Answer: (B) 4 / (9π²)
Given β = 3π, β/2 = 3π/2. We have sin(3π/2) = -1. Therefore, [sin(β/2) / (β/2)]² = [-1 / (3π/2)]² = [ -2 / (3π) ]² = 4 / (9π²). Thus, the ratio of intensity to the central peak intensity is 4 / (9π²).
Answer: (B) Statement I is correct but Statement II is incorrect
The amplitude of light at the central maximum is proportional to the slit width a, meaning its intensity is proportional to a²; hence Statement I is correct. However, the linear width of the central peak is β₀ = 2λD / a, which is inversely proportional to the slit width a, not directly proportional; hence Statement II is incorrect.
Answer: (D) Comparable to the wavelength
Diffraction (bending around edges) is noticeable when the aperture or obstacle is comparable to the wavelength; for very large openings ray optics applies.
NCERT Class 12
Answer: (A) 18 m
The Fresnel distance is given by Z_F = a²/λ. Substituting the given values: Z_F = (3 × 10⁻³ m)² / (500 × 10⁻⁹ m) = 9 × 10⁻⁶ / 5 × 10⁻⁷ = 18 m. Beyond this distance, the spreading due to diffraction becomes larger than the slit size.
Answer: (D) 4
The Fresnel distance is Z_F = a²/λ. The beam spread (width of the central diffraction maximum) is given by w = 2λD/a. Substituting D = 2Z_F = 2(a²/λ), we get w = 2λ(2a²/λ)/a = 4a. Therefore, the ratio of the diffraction spread to the slit size a is w/a = 4.
Answer: (C) one-third of the amplitude of the central maximum
Dividing the slit of width a into three equal strips of width a/3, the path difference between corresponding points of adjacent strips is (3λ/2) / 3 = λ/2. Waves from the first and second strips arrive completely out of phase and cancel each other out, leaving only the contribution from the third strip uncancelled. Thus, the resultant amplitude is roughly one-third (A₀/3) of the central peak amplitude.
Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.
In single-slit diffraction, for the n-th secondary maximum, the slit can be divided into (2n + 1) equal strips, out of which 2n cancel each other in pairs and only 1/(2n + 1) of the slit width contributes net amplitude. Thus, the effective amplitude and intensity fall off rapidly as n increases.