Q96 / 180 · JIPMER 2016 · Physics

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 $b$

  • A
    $\frac{b}{5}$
  • B
    $\frac{b}{8}$
  • Correct answer
    C
    $\frac{b}{12}$
  • D
    $\frac{b}{4}$

Solution

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] $\Delta Y=\frac{b}{12}$

Hence, the correct answer is option: 'C'.

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