Width of 1st secondary maxima =aλ⋅D
Here
a=0.2×10−3m
λ=400×10−9m
D=100×10−2=1m
Therefore, width of 1st secondary maxima
=0.2×10−3400×10−9×1
=2×10−3m=2mm
The diffraction pattern of a light of wavelength 400nm diffracting from a slit of width 0.2mm is focused on the focal plane of a convex lens of focal length 100cm. The width of the 1st secondary maxima will be :
Held on 30 Jan 2024 · Verified 6 Jul 2026.
2mm
2cm
0.02mm
0.2mm
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