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. 0.50 mm
B. 1.25 mm
C. 1.50 mm
D. 1.75 mm

Answer

B

Explanation

Given, `d=1mm = 1xx10^(-3)m, D=1m`
`lambda=500mm=500xx10^(-9)m`
Distance of nth minima from central maxima,
`x_(n)=((2n-1)lambda)/(2)(D)/(d)`
`rArr x_(3)=((2xx3-1)xx500xx10^(-9))/(2)xx(1)/(1xx10^(-3))`
`=2.5xx500xx10^(-6)=12.5xx10^(-4)m`
`rArr x_(3)=1.25mm`