Question

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 $2I_0$ and $I_0$. If fringe width is $b$

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

Answer 1

Answer: (C)

$\frac{b}{12}$

We know that, intensity, $i=4I_0{\cos }^2\frac{\varphi }{2}$
Case I $2I_0=4I_0{\cos }^2\frac{\varphi }{2}$
$\therefore \varphi =\frac{2\pi }{\lambda }\cdot \Delta x$
$\frac{\pi }{2}=\frac{2\pi }{\lambda }\cdot \frac{Y_{1}d}{D}[\because \Delta x=\frac{Y_{1}d}{D}and\varphi =\frac{\pi }{2}]$
$\Rightarrow Y_1=\frac{\lambda D}{4d}$ ...(i)
Case II $I_0=4I_0\cdot {\cos }^2\frac{\varphi }{2}$
$\therefore \varphi =\frac{2\pi }{\lambda }\cdot \Delta x$
$\frac{2\pi }{3}=\frac{2\pi }{\lambda }\cdot \frac{Y_{2}d}{D}[\because \varphi =\frac{2\pi }{3}and\Delta x=\frac{Y_{2}d}{D}]$
$\therefore Y_2=\frac{\lambda D}{3d}$ ...(ii)
$\therefore \Delta Y=Y_2-Y_1=\frac{\lambda D}{3d}-\frac{\lambda D}{4d}=\frac{\lambda D}{12d}$
$\because \frac{\lambda D}{d}=b$ [given]
$\Delta Y=\frac{b}{12}$