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 $< s_{1}Cs_{2} = \theta$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)

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

Answer 2

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

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