Question

The angular width of the central maximum in a single slit diffraction pattern is ${60}^{\circ }$. The width of the slit is $1\mu m$. 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 $50cm$ from the slits. If the observed fringe width is $1cm$, what is slit separation distance? (i.e., the distance between the centers of each slit.)

• A. $75\mu m$
• B. $100\mu m$
• C. $25\mu m$
• D. $50\mu m$

Answer 1

The correct option is C

$25\mu m$

Step1: Given data

The angular width of the central maximum $2\alpha ={60}^{\circ }$

Width of the slit$\left(a\right)$=$1\mu m$

Width of fringe$\left(\beta \right)$= $1cm$

The distance between fringe and screen$\left(D\right)$= $50cm$

Step 2:Determine the angle $\alpha$

Consider the following figure:

According to the figure,

$\begin{array}{rcl}2\alpha & =& 60\\ \alpha & =& \frac{60}{2}\\ \alpha & =& 30\end{array}$

Step 3: Compute the wavelength

We know that for single slit diffraction, $\lambda =a\sin \theta$ where $\lambda =wavelength,a=widthofslit$. So,

$$\begin{gathered} \lambda =1\times {10}^{-6}\times \sin 30 \\ \lambda =0·5\times {10}^{-6}m \end{gathered}$$

Step 4: Compute the slit separation distance

If the observed fringe width is $1cm$ then suppose $d$ is slit separation distance.

It is understood that fringe width$\left(\beta \right)=\frac{wavelength\left(\lambda \right)\times Dis\tan cebetweenslitandscreen\left(D\right)}{Dis\tan cebetweenslit\left(d\right)}$

$$\begin{gathered} d=\frac{0·5\times {10}^{-6}\times 0·5}{0·01} \\ d=25\mu m \end{gathered}$$

So, the slit separation will be $25\mu m$.

Hence, option C is the correct answer.