Answer to Question #102514 in Optics for AA

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

As per the question,

The linear dimensions of source of light =0.3mm

maximum separation between the slits (d)=?

wave length of light$(\lambda)$ =580nm

Separation between the source and the slits(D)=0.50m

We know that, $y=\dfrac{\lambda D}{d}$

$\Rightarrow 0.3\times 10^{-3}=\dfrac{580\times10^{-9}\times0.5}{d}$

$\Rightarrow d=\dfrac{580\times10^{-9}\times0.5}{0.3\times 10^{-3}}m$

$\Rightarrow d=\dfrac{580\times10^{-6}\times5}{3}m$

$d=966.66\times 10^{-6}m$

$d=0.97mm$