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 resultant intensity in a Young's double slit experiment when both slits have intensity $I_0$ is given by:
$I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

For a path difference of $\Delta x_1=\frac{\lambda }{4}$:
${\varphi }_1=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$
$\frac{{\varphi }_1}{2}=\frac{\pi }{4}=45^{\circ }$
$I_1=4I_0{\cos }^2(45^{\circ })=4I_0{\left(\frac{1}{\sqrt{2}}\right)}^2=4I_0\times \frac{1}{2}=2I_0$

For a path difference of $\Delta x_2=\frac{\lambda }{3}$:
${\varphi }_2=\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}$
$\frac{{\varphi }_2}{2}=\frac{\pi }{3}=60^{\circ }$
$I_2=4I_0{\cos }^2(60^{\circ })=4I_0{\left(\frac{1}{2}\right)}^2=4I_0\times \frac{1}{4}=I_0$

Adding the two intensities:
$I_1+I_2=2I_0+I_0=3I_0$

Therefore, the required ratio is:
$\frac{I_1+I_2}{I_0}=\frac{3I_0}{I_0}=3$