Question

Young's double slit experimental arrangement is shown in figure. If $\lambda$ is the wavelength of light used and $\angle S_{1}C{S}_2=\theta$, then the fringe width will be :

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

Answer 1

The correct option is A $\frac{\lambda }{\theta }$

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

But here, $\theta =\frac{d}{D}\Rightarrow d=D\theta$

$\therefore \beta =\frac{D\lambda }{D\theta }=\frac{\lambda }{\theta }$