Question

In a young's double slit experiment, the first maxima is observed at a fixed point $\text{P}$ on the screen. Now the screen is continuously moved away from the plane of slits. The ratio of intensity at point $\text{P}$ to the intensity at point $\text{O}$ (center of the screen) is-

• A. Remains constant
• B. Keeps on decreasing
• C. First decreases and then increases
• D. First decreases and then becomes constant

Answer 1

The correct option is B Keeps on decreasing

Let, $I_O$ and $I_P$ be the intensity at central maxima and intensity at point $\text{P}$ respectively.

$\because I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)......\left(1\right)$

Where, $\varphi =\text{Phase difference}$

At central maxima, $\varphi =0$

$\Rightarrow I_O=4I_0$

Intensity at point $\text{P}$ ;

The phase difference $\varphi$ can be written as,

$\varphi =\frac{2\pi }{\lambda }\times \delta$

Where, $\delta =\text{Path difference}$

$\delta =d\sin \theta$

$\Rightarrow \varphi =\frac{2\pi }{\lambda }\times d\sin \theta =\frac{D}{\beta }\times 2\pi \sin \theta [\because \beta =\frac{\lambda D}{d}]$

From (1) we get,

$I_P=4I_0{\cos }^2\left(\frac{D\pi \sin \theta }{\beta }\right)$

Now $\frac{I_P}{I_O}=\frac{4I_0{\cos }^2\left(\frac{D\pi \sin \theta }{\beta }\right)}{4I_0}$

$={\cos }^2\left(\frac{D\pi \sin \theta }{\beta }\right)$

As $D$ increases, ${\cos }^2$ value decreases, Hence $\frac{I_P}{I_O}$ Keeps on decreasing.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.