Question

Visible light of wavelength $550\text{nm}$ falls on a single slit and produced its second diffraction minimum at an angle of ${45}^{\circ }$ relative to the incident direction of the light. What is the width of the slit $\left(\text{in}\mu \text{m}\right)$?

• A. $2.4\mu \text{m}$
• B. $1.4\mu \text{m}$
• C. $1.56\mu \text{m}$
• D. $2.56\mu \text{m}$

Answer 1

The correct option is C $1.56\mu \text{m}$

Given:
$\lambda =550\text{nm}$
$\theta ={45}^{\circ }forn=2$

We know that the condition for $n^{th}$ minimum is given by

$d\sin \theta =n\lambda$

Where, $\text{d = width of slit}$

$\Rightarrow d\sin {45}^{\circ }=2\times 550\text{nm}$

$\Rightarrow d\times \frac{1}{\sqrt{2}}=2\times 550\text{nm}$

$=2\times 550\times \sqrt{2}\times {10}^{-9}\text{m}$

$=1555.63\times {10}^{-9}\text{m}=1.56\times {10}^{-6}\text{m}$

$=1.56\mu \text{m}$

$\therefore$ $\left(C\right)$ is the correct option.

$\text{Why this question?}$ $\text{Caution:}{\theta }_{nD}=\frac{n\lambda }{d}$ $\text{can only be applied when}\theta \text{is very small}$ $\text{but in this particular question}\theta ={45}^{\circ }$ $\text{Which is large.}$