In Young's double slit experiment the intensities at two points, for the path difference $\frac{\lambda}{4}$…

In Young's double slit experiment the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ( $\lambda=$ wavelength of light used) are $I_1$ and $I_2$respectively. If $I_0$ denotes the intensity produced by each one of the individual slits then $\frac{I_1+I_2}{I_0}$ is equal to $\left(\cos 60^{\circ}=0.5, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)$
  1. 1
  2. 2
  3. 3
  4. 4

Solution

Phase difference, $\phi=\frac{2 \pi}{\lambda} \Delta l$ For first point, $\phi_1=\frac{2 \pi}{\lambda}\left(\frac{\lambda}{4}\right)$ $\begin{aligned} \therefore \quad \phi_1 & =\frac{\pi^c}{2} \\ \mathrm{I}_1 & =2 \mathrm{I}_0\left(1+\cos \phi_1\right) \\ \therefore \quad \mathrm{I}_1 & =2 \mathrm{I}_0 \quad \ldots .\left[\because \cos \phi_1=\cos (\pi / 2)=0\right] \end{aligned}$ Similarly, for second point, $\phi_2=\frac{2 \pi^e}{3}$ $\begin{array}{ll} \therefore \quad \mathrm{I}_2 & =2 \mathrm{I}_0\left(1+\cos \phi_2\right) \\ \therefore \quad \mathrm{I}_2 & =2 \mathrm{I}_0\left(1-\frac{1}{2}\right) \quad \ldots . .\left[\because \cos \left(\frac{2 \pi}{3}\right)=\frac{-1}{2}\right] \\ \therefore \quad \mathrm{I}_2 & =\mathrm{I}_0 \end{array}$ Hence, $\frac{I_1+I_2}{I_0}=\frac{2 I_0+I_0}{I_6}=3$

Asked in: MHT CET 2023 (12 May Shift 1)

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