Question

The intensity of each of the two slits in Young's double slit experiment is $I_0$. Calculate the minimum separation between the two points on the screen where intensities are $2I_0$ and $I_0$. Given, the fringe width equal to $\beta$.

• A. $\frac{\beta }{4}$
• B. $\frac{\beta }{3}$
• C. $\frac{\beta }{12}$
• D. $\beta$

Answer 1

The correct option is C $\frac{\beta }{12}$

If $I_0$ is the intensity of each slit, then Intensity at a point on the screen is given by,

$I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

Let, $I_1=2I_0$

$\Rightarrow 2I_0=4I_0{\cos }^2\left(\frac{{\varphi }_1}{2}\right)$
$\Rightarrow {\varphi }_1=\frac{\pi }{2}$

Let, $I_2=I_0$
$\Rightarrow I_0=4I_0{\cos }^2\left(\frac{{\varphi }_2}{2}\right)$
$\Rightarrow {\varphi }_2=\frac{2\pi }{3}$

As, $\varphi =\left(\frac{2\pi }{\lambda }\right)\Delta x\text{and}\Delta x=\frac{yd}{D}$

$\Rightarrow \varphi =\left(\frac{2\pi }{\lambda }\right)\left(\frac{yd}{D}\right)$

Here, $\beta =\frac{\lambda D}{d}$

$\Rightarrow \varphi =\frac{2\pi y}{\beta }$

For, ${\varphi }_1=\frac{\pi }{2}\Rightarrow y_1=\frac{\beta }{4}$

For, ${\varphi }_2=\frac{2\pi }{3}\Rightarrow y_2=\frac{\beta }{3}$

The distance between these two points,
$\Delta y=y_2-y_1=\frac{\beta }{12}$

Hence, option $\left(C\right)$ is correct.