In a YDSE, `lambda=4000 Å`, fringes observed have a width `beta`. The light illuminating the set-up now has `lambda=6000 Å` and the separation between the interfering sources is halved. What is the ratio of the distance between the screen and the interfering sources before and now if the fringe and now if the fringe width ramains unaltered

Options

A. `1//3`
B. `3//1`
C. `3//4`
D. `2//3`

Answer

B

Explanation

`beta=(D_(1)lambda_(1))/d, beta=(D_(2)lambda_(2))/(d//2)`
`(D_(1)lambda_(1))/d=(2D_(2)lambda_(2))/d implies D_(1)/D_(2)=(2lambda_(2))/lambda_(1)=(2xx6000)/4000=3/1`