In a double slit experiment, the two slits are `1mm` apart and the screen is placed `1m` away. A monochromatic light of wavelength `500nm` is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single-slit pattern?

Options

A. `0.5` mm
B. `0.02` mm
C. `0.2` mm
D. `0.1` mm

Answer

C

Explanation

For double slit experiment,
`d=1mm=1xx10^(-3)m,D=1m,lambda=500xx10^(-9)m`
Fringe width `beta=(Dlambda)/(d)`
Width of central maxima in a single slit `=(2lambdaD)/(a)`
As per question,
width of central maxima of single slit pattern = width of `10` maxima of double slit pattern
`(2lambdaD)/(a)=10((lambdaD)/(d))`
`impliesa=(2d)/(10)=(2xx10^(-3))/(10)=0.2xx10^(-3)m=0.2mm`