Question

In Young's slit experiment, interference fringes are observed on a screen, kept at D from the slits. If the screen is moved towards the slits by 5 x 10⁻² m the change in fringe width is found to be 3×l0⁻⁵ m. If the separation between the slits is 10-3in, calculate the wavelength of the light used.

Answer 1

it is given that, change in distance between screen and slits , ∆D= 5 × 10^-2m

change in fringe width , $\Delta\beta$ = 3 × 10^-5 m

separation between slits , d = 10^-3 in = 2.54 × 10^-3 cm = 2.54 × 10^-5 m

use formula, $\beta=\frac{\lambda D}{d}$

differentiating both sides,

$d\beta=\frac{\lambda dD}{d}$

or, $\Delta\beta=\frac{\lambda\Delta D}{d}$

or, $\lambda=\frac{\Delta\beta d}{\Delta D}$

= 3 × 10^-5 × 2.54 × 10^-5/5 × 10^-2

= (7.62 × 10^-10)/5 × 10^-2

= 1.524 × 10^-8 m

hence, wavelength of light used = 125.4 A°