Physics - Wave Optics Question with Solution | TestHub

If$\delta$is the phase difference between the interfering waves at point$P,$then the intensity at point$P$is given by

cos2⁡δ2

Given$I=\frac{I_{max}}{2}.$Hence

${cos}^2\left(\frac{\delta }{2}\right)=\frac{1}{2},$which gives$\left(\frac{\delta }{2}\right)=\frac{\pi }{4}$

Or$\delta =\frac{\pi }{2}$

The angular separation$\theta$between points$P$and$O$is given by$\tan \theta =\frac{y}{D}$. Since$\theta \simeq \sin \theta .$Hence

$\sin \theta =\frac{y}{D}$...... (i)

If$\beta$is the fringe width, then

$\frac{y}{\beta }=\frac{\pi /2}{2\pi }=\frac{1}{4}$...... (ii)

This is because the phase difference$\delta$between two consecutive maxima is$2\pi$. Now$\beta =\frac{\lambda D}{d}.$using this in Equation. (ii), we get

$\frac{yd}{\lambda d}=\frac{1}{4}$

Or$\frac{y}{D}=\frac{\lambda }{4d}$.....(iii)

Using Equation. (i) in Equation. (iii), we have

$\sin \theta =\frac{\lambda }{4d} \mathrm{o}\mathrm{r} \theta ={\sin }^{-1}\left(\frac{\lambda }{4d}\right)$