Question

Light of wavelength $550\text{nm}$ passes through a narrow single slit of width $2.00\mu \text{m}$ and produces a diffraction pattern on the screen placed few meters away from the slit. Calculate the intensity relative to the central maximum at a point halfway between these two minima.

$\text{Take}:{\sin }^{-1}\left(0.275\right)={16}^{\circ }$
${\sin }^{-1}\left(0.550\right)={33.4}^{\circ },$
$\sin \left({24.7}^{\circ }\right)=0.417,$
$\sin \left(4.77\text{rad}\right)=-0.998$

• A. $0.4$
• B. $0.04$
• C. $4$
• D. $0.2$

Answer 1

The correct option is B $0.04$

Given :

$\lambda =550\text{nm}=550\times {10}^{-9}\text{m}$
$b=2\mu \text{m}=2\times {10}^{-6}\text{m}$

Using:
$b\sin \theta =n\lambda \text{[For minima]}\text{First two minima are at n=1, 2}$

${\theta }_1={\sin }^{-1}\left(\frac{\lambda }{b}\right)$

${\theta }_1={\sin }^{-1}\left(\frac{550\times {10}^{-9}}{2\times {10}^{-6}}\right)={\sin }^{-1}\left(0.275\right)$

$\therefore {\theta }_1={16}^{\circ }$

similarly,
${\theta }_2={\sin }^{-1}\left(\frac{2\times 550\times {10}^{-9}}{2\times {10}^{-6}}\right)={\sin }^{-1}\left(0.550\right)$

$\therefore {\theta }_2={33.4}^{\circ }$

Therefore, halfway between ${\theta }_1$ and ${\theta }_2$,

$\theta =\frac{{\theta }_1+{\theta }_2}{2}=\frac{16+33.4}{2}={24.7}^{\circ }$

$\therefore$ angular separation will be,

$\beta =\frac{\pi b\sin \theta }{\lambda }=\frac{\pi \times 2\times {10}^{-6}\times \sin {24.7}^{\circ }}{550\times {10}^{-9}}$

$\Rightarrow \beta =4.77\text{rad}$

Now, intensity at a point on a screen,

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

$\Rightarrow \frac{I}{I_0}={\left(\frac{\sin \left(4.77\right)}{4.77}\right)}^2$

$\therefore \frac{I}{I_0}={\left(\frac{-0.998}{4.77}\right)}^2=0.043$

So, option $\left(\text{B}\right)$ is the correct answer.