Question

an extended source of light whose linear dimension is 0.3nm is used in young's double slit experiment .calculate the maximum seperation between the slits for which interference pattern will be observable if the wavelength of light is 580nm and seperation between the source and the slits is 0.50m.

Answer 1

In YDSE condition for maximum is

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

$d =\lambda/sin\theta$

$d=λ∗ (D/x)$

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