Diffraction (due to a single slit, width of central maximum ) NEET Questions

Physics - Optics

Diffraction (due to a single slit, width of central maximum ) MCQ Questions


13.
The width of each SECONDARY MAXIMUM in a single-slit diffraction pattern is:
A.
3λD/(2a)
B.
2λD/a (same as the central maximum width)
C.
λD/(2a)
D.
λD/a (half the width of the central maximum = W/2)
😑 View Answer Rough Work Error
ANSWER :
D. λD/a (half the width of the central maximum = W/2)
14.
A student performs a single-slit diffraction experiment with red light (λ = 700 nm) and then with blue light (λ = 400 nm), keeping slit width and screen distance constant. How do the central maximum widths compare?
A.
4 : 7 (blue is wider because it has higher frequency)
B.
Red central maximum width : Blue central maximum width = 7 : 4 (since W ∝ λ, W_red/W_blue = 700/400 = 7/4)
C.
1 : 4 (blue has 4 times the frequency so 4 times the diffraction)
D.
1 : 1 (colour does not affect diffraction width)
😑 View Answer Rough Work Error
ANSWER :
B. Red central maximum width : Blue central maximum width = 7 : 4 (since W ∝ λ, W_red/W_blue = 700/400 = 7/4)
15.
The ANGULAR HALF-WIDTH of the central maximum equals the angle at which the first minimum occurs. This angle is given by:
A.
sinθ₁ = a/λ (inverse relationship)
B.
sinθ₁ = λ/a (exactly), or θ₁ ≈ λ/a (for small angles in radians)
C.
sinθ₁ = λ/(2a)
D.
sinθ₁ = 2λ/a
😑 View Answer Rough Work Error
ANSWER :
B. sinθ₁ = λ/a (exactly), or θ₁ ≈ λ/a (for small angles in radians)
16.
The intensity distribution in single-slit diffraction follows the formula:
A.
I(θ) = I₀ [cos(β/2)/(β/2)]²
B.
I(θ) = I₀ [sin(β/2)/(β/2)]² where β = (2π/λ) a sinθ is the phase difference across the full slit width
C.
I(θ) = I₀ sin²(β/2)
D.
I(θ) = I₀ cos²(β/2)
😑 View Answer Rough Work Error
ANSWER :
B. I(θ) = I₀ [sin(β/2)/(β/2)]² where β = (2π/λ) a sinθ is the phase difference across the full slit width
17.
The intensity of the FIRST SECONDARY MAXIMUM in single-slit diffraction relative to the central maximum is approximately:
A.
4.7% (≈ I₀/22) — the first secondary maximum occurs near a sinθ = 3λ/2, giving (sinα/α)² ≈ (sin(3π/2)/(3π/2))² = (1/(3π/2))² ≈ 4/(9π²) ≈ 4.5%
B.
25% (I₀/4)
C.
11% (I₀/9)
D.
50% (I₀/2)
😑 View Answer Rough Work Error
ANSWER :
A. 4.7% (≈ I₀/22) — the first secondary maximum occurs near a sinθ = 3λ/2, giving (sinα/α)² ≈ (sin(3π/2)/(3π/2))² = (1/(3π/2))² ≈ 4/(9π²) ≈ 4.5%
18.
In the single-slit diffraction intensity pattern, as we move from the centre outward:
A.
Intensity remains constant — all maxima have the same brightness
B.
Intensity alternates equally between bright and dark — like double-slit interference
C.
Intensity increases — outer fringes are brighter due to more diffraction
D.
Intensity decreases — the central maximum is the brightest; secondary maxima progressively diminish in intensity
😑 View Answer Rough Work Error
ANSWER :
D. Intensity decreases — the central maximum is the brightest; secondary maxima progressively diminish in intensity