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Q. In a Young's double slit experiment, let $S_1$ and $S_2$ be the two slits, and $C$ be the centre of the screen. If $\angle S_1C S_2 = \theta$ and $\lambda$ is the wavelength, the fringe width will be

Wave Optics

Solution:

Fringe width, $\beta = \frac{\lambda D}{d}$ and
$\theta = \frac{d}{D} $
$\therefore \beta =\frac{ \lambda }{\theta}$