Question

In a Young's double-slit experiment, let $S_1$ and $S_2$ be the two slits, and $C$ be the center of the screen. If $\angle S_{1}C{S}_2=\theta$ and $\lambda$ is the wavelength, the fringe width will be

• A. $\frac{\lambda }{\theta }$
• B. $\lambda \theta$
• C. $\frac{2\lambda }{\theta }$
• D. $\frac{\lambda }{2\theta }$

Answer 1

Answer: (A)

$\frac{\lambda }{\theta }$

The fringe width for the pattern is given by:

$\beta =\frac{\lambda D}{d}$,

From the diagram, for larger value of the distance of the screem from the slit, (d<<D)

$\theta =\frac{d}{D}$

So, the fringe width can be written as:
$\therefore \beta =\frac{\lambda }{\theta }$