Question

In a Fresnel biprism experiment, if prism angle $\alpha =1^{\circ };\mu =\mathrm{1.54.}$ The distance between screen $\&$ prism $b=0.7\text{m}$, the distance between prism $\&$ source $a=0.3\text{m},\lambda =180\pi \text{nm}.$ Find the value of $\beta$ (fringe width).

• A. ${10}^{-4}\text{m}$
• B. ${10}^{-3}\text{mm}$
• C. ${10}^{-4}\times \pi \text{m}$
• D. $\pi \times {10}^{-3}\text{m}$

Answer 1

The correct option is A ${10}^{-4}\text{m}$

Given:
$\mu =1.54;\alpha =1^{\circ }=\frac{\pi }{{180}^{\circ }};a=0.3\text{m}$
$b=0.7\text{m};D=0.7\text{m};\lambda =180\pi \text{nm}$


$d=2a(\mu -1)\alpha$

The fringe width,

$\beta =\frac{D\lambda }{d}=\frac{(a+b)\lambda }{2a(\mu -1)\alpha }$

$=\frac{(0.3+0.7)\times 180\pi \times {10}^{-9}}{2\times 0.3(1.54-1)\times \frac{\pi }{180}}$

$\beta =\frac{324\times {10}^{-7}}{0.324}={10}^{-4}\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question? To understand the concept of fringe in a Fresnel biprism experiment.