Consider a single-slit diffraction pattern for a slit of width d. It is observed that for a light of wavelength 400 nanometers, the angle between the first minimum and the central maximum is 4 × 10−3 radians. The value of d is
A. 1 × 10−5 m
B. 5 × 10−5 m
C. 1 × 10−4 m
D. 2 × 10−4 m
E. 1 × 10−3 m
(GR9677 #57)
Single slit diffraction: d sin θ = mλ
λ = 400 nm = 4 × 10−7 m
θ = 4 × 10−3 radians
sin θ ≈ θ for small θ
dθ = mλ
d = mλ / θ = 1 × (4 × 10−7) / (4 × 10−3) = 10−4 m
Answer: C
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