Solveeit Logo
Login

Question

Physics Question on Optics

The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1st secondary maxima will be :

A

2 mm

B

2 cm

C

0.02 mm

D

0.2 mm

Answer

2 mm

Explanation

Solution

The width of the first secondary maxima for single-slit diffraction is given by:

Width of 1st secondary maxima = λDa\frac{\lambda D}{a}

where λ=400×10−9 m\lambda = 400 \times 10^{-9} \, \text{m}, a=0.2×10−3 ma = 0.2 \times 10^{-3} \, \text{m}, and D=100 cm=1 mD = 100 \, \text{cm} = 1 \, \text{m}. Substitute the values:

Width of 1st secondary maxima = 400×10−9×10.2×10−3=2 mm\frac{400 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 2 \, \text{mm}