In a Young's double slit experiment, the slit separation is `1mm` and the screen is `1m` from the slit. For a monochromatic light of wavelength `500nm`, the distance of 3rd minima from the central maxima is

Options

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

Answer

D

Explanation

`d=1mm=10^(-3), D=1m, lambda=500 nm=500xx10^(-9) m`
width of central maximum in single slit pattern
`=(2D lambda)/a, a`: slit width
Fringe width in double slit experiment
`beta=(Dlambda)/d`
So required condiction `implies (10 D lambda)/d=(2D lambda)/a`
`a=d/5=1/5xx10^(-3) m=0.2xx10^(-3) m`
`=0.2 mm^(3)`