In Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{6}$…

In Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{6}$ ( 1 being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity. $\frac{\mathrm{I}}{\mathrm{I}_0}$, is equal to
  1. $\frac{1}{\sqrt{2}}$
  2. $\frac{\sqrt{3}}{2}$
  3. $\frac{1}{2}$
  4. $\frac{3}{4}$

Solution

In YDSE, $\Delta x=\frac{\lambda}{6}, \mathrm{I}_1=\mathrm{I}_2$ $\therefore$ Phase difference, $\Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}$ $\begin{aligned} & \therefore \mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi \\ & =\mathrm{I}_1+\mathrm{I}_1+2 \mathrm{I}_1 \cos \frac{\pi}{3}=3 \mathrm{I}_1 \\ & \mathrm{I}_0=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2=\left(2 \sqrt{\mathrm{I}_1}\right)^2=4 \mathrm{I}_1 \\ & \therefore \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{3 \mathrm{I}_1}{4 \mathrm{I}_1}=\frac{3}{4} \end{aligned}$

Asked in: AP EAMCET 2024 (19 May Shift 2)

Practice more Wave Optics questions on Aicharya