Two slits at a distance of `1mm` are illuminated by a light of wavelength `6.5xx10^-7m`. The interference fringes are observed on a screen placed at a distance of `1m`. The distance between third dark fringe and fifth bright fringe will be

Options

A. 0.65 mm
B. 1.30 mm
C. 1.625 mm
D. 0.975 mm

Answer

C

Explanation

As Dgtgtd and `lambda`ltltd
Hence we can use.
`beta=(lambdaD)/d`
so distance between `5^(th)` bright fringe and `3^(rd)` dark fringe `=5xx beta-(2 beta+beta//2)`
`=5/2 beta=5/2 xx(6.5xx10^(-7)xx1)/(10^(-3))`
=1.625 mm