Question

The fringe width $\beta$ can be expressed as

• A. $\beta ={\beta }_0\sin \omega t$
• B. $\beta ={\beta }_0\cos \omega t$
• C. $\beta ={\beta }_0\sin 2\omega t$
• D. none of these

Answer 1

The correct option is D none of these

At any time, path difference between the lights from slits $S_1$ and $S_2$ reaching a point P ($y$ distance away from central maxima) on the screen is $y\frac{d}{D}$

Let for maxima, $\frac{yd}{D}=n\lambda ⟹y=\frac{n\lambda D}{d}$

Thus fringe width, $\beta =\frac{\lambda D}{d}$ for all time