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 Young's double slit experiment for maximum

$d\sin \alpha=n\lambda$

where d is the separation between the slits.

For small $\alpha$

$\sin\alpha\approx \tan\alpha \approx\alpha$

We have (for $n=1$ )

$x\cdot\sin \alpha=\lambda$

$\sin\alpha\approx \frac{d}{L}$

$x\cdot \frac{d}{L}=\lambda \to d=\frac{L\cdot \lambda}{x}=\frac{0.5\cdot 580\cdot 10^{-9}}{0.3\cdot 10^{-3}}=967\cdot 10^{-6}m=0.967 mm.$