Light of wavelength 6000A falls on a single slit of width 0.1mm. The second minimum will be formed for the angle of diffraction of

Options

A. 0.08rad
B. 0.06rad
C. 0.12rad
D. 0.012rad

Answer

D

Explanation

Given single slit of width d=0.1mm
`d=0.1 times 10^-3 m`
or `d=1 times 10^-4 m`
Light of wavelength `lamda=6000A`
or `lamda=6 times 10^-7 m`
The angle of diffraction
`theta= (n lamda)/d`
`theta= (2 times 6 times10^-7)/(1 times 10^-4)`
`theta=12 times 10^-3`
`theta=0.02 rad`