In Young's double slit experiment, the wavelength of light used is ' $\lambda$ '. The intensity at a point…

In Young's double slit experiment, the wavelength of light used is ' $\lambda$ '. The intensity at a point is ' $I$ ' where path difference is $\left(\frac{\lambda}{4}\right)$. If $I_0$ denotes the maximum intensity, then the ratio $\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)$ is $\left(\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)$
  1. $\frac{1}{\sqrt{2}}$
  2. $\frac{1}{2}$
  3. $\frac{3}{4}$
  4. $\frac{\sqrt{3}}{2}$

Solution

Phase difference, $\Delta \phi=\left(\frac{2 \pi}{\lambda}\right) \Delta l$ For path difference $\frac{\lambda}{4}$, Phase difference $\Delta \phi=\frac{\pi}{2}$ Using, $\mathrm{I}=\mathrm{I}_0 \cos ^2 \frac{\phi}{2}$ $\begin{array}{ll} \therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\cos ^2 \frac{\phi}{2}=\cos ^2\left(\frac{\pi}{4}\right) \\ \therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{1}{2} \end{array}$

Asked in: MHT CET 2023 (09 May Shift 2)

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