Answer to Question #102459 in Optics for Abhishek

Question

An extended source of light whose linear dimension is 0.3 mm is used in Young’s double slit experiment. Calculate the maximum separation between the slits for which interference pattern will be observable if the wavelength of light is 580 nm and separation between the source and the slits is 0.50 m.

Answer 1

In YDSE condition for maximum is

$dsin\theta=\lambda$ , where d is the separation between the slits

$d$ $=\lambda/sin\theta$

$d=\lambda*$ $D/x$

$d=0.58*10^{-6}*0.5/0.0003=0.97*10^{-3}m = 0.97 mm$