In Young's experiment, interference fringes are obtained on a screen placed at some distance from the slits. When the screen is moved towards the slits by `5xx10^(-2)m`, the fringe width is changed by `3.5xx10^(-5)m`. If the separation between the slits is 1 mm, then th wavelength of light used is

Options

A. 4000 Å
B. 5000 Å
C. 6000 Å
D. 7000 Å

Answer

D

Explanation

`X_(1)-X_(2)=(lamda[D_(1)-D_(2)])/(d)`
`therefore 3.5 xx10^(-5)=(lamda[5xx10^(-2)])/(10^(-3))`
`therefore lamda=(3.5xx10^(-6))/(5)=7000` Å