Physics - Wave Optics Question with Solution | TestHub

The distance between two successive bright fringes is the fringe width, $\beta =\frac{\lambda D}{d}$.

The distance of the given point from the centre of the bright fringe is $y=\frac{\beta }{4}$.

The path difference at this point is $\Delta x=\frac{yd}{D}=\frac{\beta d}{4D}=\frac{\lambda }{4}$.

The corresponding phase difference is $\Delta \varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }\left(\frac{\lambda }{4}\right)=\frac{\pi }{2}$.

The intensity at any point in the interference pattern is given by $I=I_{max}{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$.

At the centre of a bright fringe, $\Delta \varphi =0$, so the intensity is $I_0=I_{max}{\cos }^2(0)=I_{max}$.

At the given point, the intensity is $I=I_{max}{\cos }^2\left(\frac{\pi /2}{2}\right)=I_{max}{\cos }^2\left(\frac{\pi }{4}\right)=I_{max}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_{max}}{2}$.

The required ratio of the intensity at the centre of the bright fringe to the intensity at the given point is:

$\frac{I_0}{I}=\frac{I_{max}}{\frac{I_{max}}{2}}=2$

Answer: $2$