Question

The angular width of the central maximum in a single slit diffraction pattern is ${60}^{\circ }$. The width of the slit is $1\mu \text{m}$. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young's Fringes can be observed on a screen placed at a distance $50\text{cm}$ from the slits. If the observed fringe width is $1\text{cm}$. What is slit separation distance? $\left(\text{i.e. distance between the centre of each slit}\right)$

Answer 1

In single slit diffraction,

The angular width of central maximum is given by $2\theta$

Where, $\theta -\text{angular position of first minimum}$

Given that, $2\theta ={60}^{\circ }$
$\Rightarrow \theta ={30}^{\circ }$

The condition or first minimum is given by,

$a\sin \theta =\lambda$

$\Rightarrow 1\times {10}^{-6}\sin {30}^{\circ }=\lambda$

$\Rightarrow \lambda =0.5\times {10}^{-6}\text{m}=5\times {10}^{-7}\text{m}$



In $YDSE$,

Fringe width, $\beta =\frac{\lambda D}{d}$

Where $\text{d = distance between the slits}$

$\Rightarrow 1\times {10}^{-2}=\frac{5\times {10}^{-7}\times 50\times {10}^{-2}}{d}$

$\Rightarrow d=25\times {10}^{-6}\text{m}=25\mu \text{m}$

$\begin{array}{c}\text{Why this question?}\\ \text{Asked in JEE mains 2018.}\end{array}$