Question

In a single slit diffraction experiment, light of wavelength $500\text{nm}$ passes through a slit of width $2\mu \text{m}$. If the intensity of the central maximum is $40{\text{Wm}}^{-2}$, then the intensity of a point on a screen at an angular position of $\theta ={30}^{\circ }$ is close to

• A. $0$
• B. $20{\text{Wm}}^{-2}$
• C. $1{\text{Wm}}^{-2}$
• D. $2{\text{Wm}}^{-2}$

Answer 1

The correct option is A $0$

Given:
$\lambda =500\text{nm}$
$a=2\mu \text{m}$

The intensity at a point on the screen is given by

$I=I_0{\left(\frac{\sin \beta }{\beta }\right)}^2$

Where,
$I_0-\text{Intensity at the central maximum}$

$\beta =\frac{\pi a\sin \theta }{\lambda }$

Here, $\theta ={30}^{\circ }$

$\Rightarrow \beta =\frac{\pi \times 2\times {10}^{-6}\times \left(\frac{1}{2}\right)}{500\times {10}^{-9}}$

$\Rightarrow \beta =2\pi$

$\Rightarrow I=I_0{\left(\frac{\sin 2\pi }{2\pi }\right)}^2$

$\Rightarrow I=0$

Hence, option $\left(\text{A}\right)$ is correct.

Alternate Method: The condition for minimum in diffraction pattern is given by, $a\sin \theta =n\lambda$ $\Rightarrow 2\times {10}^{-6}\times \left(\frac{1}{2}\right)=n\times 500\times {10}^{-9}$ $\Rightarrow n=2$ $\Rightarrow$ Second minimum will be formed at the given angular position, hence, the intensity will be zero.