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

Here, `d = 1 mm, D = 1 m, lambda = 500 nm`
In double slit experiment,
angular width `= (beta)/(D)` as `beta = (lambda D)/(d)`
`:. (beta)/(D) = (lambda D)/((d)/(D)) = (lambda)/(d)`
In single slit experiment,
angular width `= (2 lambda)/(d')`
According to Question,
`(10 lambda)/(d) = (2 lambda)/(d') = or d' = 0.2 d`
`d' = 0.2 mm`