Question

In a double slit experiment, the separation between the slits is $d$ and distance of the screen from slits is $D$. If the wavelength of light used is $\lambda$ and $I$ is the intensity of central bright fringe, then intensity at distance $x$ from central maximum is

• A. $I{\cos }^2\left(\frac{{\pi }^{2}xd}{\lambda D}\right)$
• B. $I^2{\sin }^2\left(\frac{\pi xd}{2\lambda D}\right)$
• C. $I{\cos }^2\left(\frac{\pi xd}{\lambda D}\right)$
• D. $I{\sin }^2\left(\frac{\pi xd}{\lambda D}\right)$

Answer 1

The correct option is C $I{\cos }^2\left(\frac{\pi xd}{\lambda D}\right)$

For a path difference $\Delta$, the phase difference (in radians) is

$\delta =\frac{2\pi }{\lambda }\Delta$

We know,

Path difference, $\Delta =\frac{x_{n}d}{D}$
where $x_n$ is the distance from the central maxima
$d$ is the separation between the slits
$D$ is the distance of the screen from slits

Here,
Given $x_n=x$
Path difference, $\Delta =\frac{xd}{D}$

$\Rightarrow \delta =\frac{2\pi }{\lambda }\frac{xd}{D}=\frac{2\pi xd}{D\lambda }$

Intensity due to interference is given by

$I=I_1+I_2+2\sqrt{I_1{I}_2}Cos\delta$

At central maximum, $\delta =0$,

Given, $I_{center}=I$

Let the intensity of light from each slit be $I_1=I_2=I_s$

$\Rightarrow I=I_s+I_s+2\sqrt{I_s{I}_s}Cos0{}^{\circ }$

$\Rightarrow I=4I_s$

Now, intensity at a distance $x$ is given by

$I_x=I_s+I_s+2\sqrt{I_s{I}_s}Cos\delta$

$I_x=2I_s+2I_{s}Cos\delta$

$I_x=\frac{I}{2}+\frac{I}{2}Cos\left(\frac{2\pi xd}{D\lambda }\right)$

$I_x=\frac{I}{2}(1+Cos\left(\frac{2\pi xd}{D\lambda }\right))$

Using

$Cos2\theta =2Co{s}^2\theta -1$

$\Rightarrow \frac{1+Cos2\theta }{2}=Co{s}^2\theta$

We have,
$I_x=ICo{s}^2\left(\frac{\pi xd}{D\lambda }\right)$

Hence, the correct answer is OPTION C.