In Young's double slit experiment, the resultant intensity of light at a point on the screen is 'I' when the…

In Young's double slit experiment, the resultant intensity of light at a point on the screen is 'I' when the path difference is $\lambda^{\prime}$ '. When the path difference is $\frac{\lambda}{4}$, the intensity at a point will be $\left(\lambda=\right.$ wavelength of light, $\cos 180^{\circ}=-1, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$ )
  1. Zero
  2. I
  3. $\frac{\mathrm{I}}{2}$
  4. $\frac{\mathrm{I}}{4}$

Solution

$\begin{aligned} & \mathrm{I}=2 \mathrm{I}_{\mathrm{o}}(1+\cos \theta) \\ & \phi=\frac{2 \pi}{\lambda}(\text { path as difference })\end{aligned}$ $\begin{aligned} \mathrm{I} & =2 \mathrm{Z}_{\mathrm{o}}\left(1+\cos \left(\frac{2 \mathrm{z}}{\lambda} \times \lambda\right)\right) \\ \mathrm{I} & =4 \mathrm{I}_{\mathrm{o}} \\ \mathrm{I}^{\prime} & =2 \mathrm{I}_{\mathrm{o}}\left(1+\cos \left(\frac{2 \mathrm{z}}{\lambda} \times \frac{\lambda}{4}\right)\right) \\ \mathrm{I}^{\prime} & =2 \mathrm{I}_{\mathrm{o}}\left(1+\cos \left(\frac{2 \mathrm{z}}{\lambda}\right)\right. \\ & =2 \mathrm{I}_{\mathrm{o}} \end{aligned}$ So,$I^{\prime}=I / 2$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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