In a single slit diffraction with `lambda = 500` nm and a lens of diameter 0.1 mm then width of central maxima, obtain on screen at a distance of l m will be ......

Options

A. 5 mm
B. 1 mm
C. 10 mm
D. `2.5` mm

Answer

C

Explanation

The width of central maximum means distance between first minimum on both side of it.
`2 sin v_(1)=(2lambdaD)/(2)`
`=(2xx5xx10^(-5)xx100)/(10^(-2))=10xx10^(-1) cm=10 mm`