Physics - Wave Optics Question with Solution | TestHub

The phase difference $\varphi$ is related to the path difference $\Delta x$ by the formula:

$\varphi =\frac{2\pi }{\lambda }\Delta x$

The intensity at any point on the screen is given by:

$I=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

For point P, the path difference is $\Delta x_P=0$.

${\varphi }_P=0$

$I_P=I_{max}{\cos }^2(0)=I_{max}$

For point Q, the path difference is $\Delta x_Q=\frac{\lambda }{4}$.

${\varphi }_Q=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$

$I_Q=I_{max}{\cos }^2\left(\frac{\pi /2}{2}\right)=I_{max}{\cos }^2\left(\frac{\pi }{4}\right)$

$I_Q=I_{max}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_{max}}{2}$

The ratio of intensities at point P to that at Q is:

$\frac{I_P}{I_Q}=\frac{I_{max}}{\frac{I_{max}}{2}}=\frac{2}{1}$

The ratio is $2:1$.

Answer: $2:1$