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The intensity of each of the two slits in Young's double slit experiment is $I_0$. Calculate the minimum separation between the points on the screen, where intensities are $2 I_0$ and $I_0$. If fringe width is $b$
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The correct answer is:
$\frac{b}{12}$
We know that, intensity, $I=4 I_0 \cos ^2 \frac{\phi}{2}$
$\begin{aligned} & \text { Case I } 2 I_0=4 I_0 \cos ^2 \frac{\phi}{2} \\ & \therefore \quad \phi=\frac{2 \pi}{\lambda} \cdot \Delta x \\ & \frac{\pi}{2}=\frac{2 \pi}{\lambda} \cdot \frac{Y_1 d}{D}\left[\because \Delta x=\frac{Y_1 d}{D} \text { and } \phi=\frac{\pi}{2}\right] \\ & \Rightarrow \quad Y_1=\frac{\lambda D}{4 d}......(i) \\ & \end{aligned}$
Case II $I_0=4 I_0 \cdot \cos ^2 \frac{\phi}{2}$
$\therefore \quad \phi=\frac{2 \pi}{\lambda} \cdot \Delta x$
$\frac{2 \pi}{3}=\frac{2 \pi}{\lambda} \cdot \frac{Y_2 d}{D} \quad\left[\because \phi=\frac{2 \pi}{3}\right.$ and $\left.\Delta x \frac{y_2 d}{D}\right]$
$\therefore \quad Y_2=\frac{\lambda D}{3 d}$.....(ii)
$\therefore \quad \Delta Y=Y_2-Y_1=\frac{\lambda D}{3 d}-\frac{\lambda D}{4 d}=\frac{\lambda D}{12 d}$
$\because \quad \frac{\lambda D}{d}=b$
[given]
$\Delta Y=\frac{b}{12}$
$\begin{aligned} & \text { Case I } 2 I_0=4 I_0 \cos ^2 \frac{\phi}{2} \\ & \therefore \quad \phi=\frac{2 \pi}{\lambda} \cdot \Delta x \\ & \frac{\pi}{2}=\frac{2 \pi}{\lambda} \cdot \frac{Y_1 d}{D}\left[\because \Delta x=\frac{Y_1 d}{D} \text { and } \phi=\frac{\pi}{2}\right] \\ & \Rightarrow \quad Y_1=\frac{\lambda D}{4 d}......(i) \\ & \end{aligned}$
Case II $I_0=4 I_0 \cdot \cos ^2 \frac{\phi}{2}$
$\therefore \quad \phi=\frac{2 \pi}{\lambda} \cdot \Delta x$
$\frac{2 \pi}{3}=\frac{2 \pi}{\lambda} \cdot \frac{Y_2 d}{D} \quad\left[\because \phi=\frac{2 \pi}{3}\right.$ and $\left.\Delta x \frac{y_2 d}{D}\right]$
$\therefore \quad Y_2=\frac{\lambda D}{3 d}$.....(ii)
$\therefore \quad \Delta Y=Y_2-Y_1=\frac{\lambda D}{3 d}-\frac{\lambda D}{4 d}=\frac{\lambda D}{12 d}$
$\because \quad \frac{\lambda D}{d}=b$
[given]
$\Delta Y=\frac{b}{12}$
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