Question

Light of wavelength `6328Å` is incident normally on slit having a width of `0.2mm`The width of the central maximum measured form minimum to minimum of diffraction pattens on a screen `9.0meter`away will be about-
A. `0.36^(@)`
B. `0.18^(@)`
C. `0.72^(@)`
D. `0.09^(@)`

Answer 1

Correct Answer - 1
Silit width =`a=0.2mm,beta=(2lambdaD)/(a)`
Angular width `W_(theta)=(beta)/(D)=(2lambda)/(a) rArr theta =(2xx6328)/(0.2)=0.36^(@)`