Question

In a Young's double slit experiment, light of $500\text{nm}$ is used to produce an interference pattern. When the distance between the slits is $0.05\text{mm}$, the angular width (in degree) of the fringes formed on the distance screen is close to-

• A. ${0.17}^{\circ }$
• B. ${0.57}^{\circ }$
• C. ${1.7}^{\circ }$
• D. ${0.07}^{\circ }$

Answer 1

The correct option is B ${0.57}^{\circ }$

Given, $\lambda =500\text{nm};d=0.05\text{mm}$

Angular width of the fringe formed,

$\theta =\frac{\beta }{D}=\frac{\lambda }{d}=\frac{500\times {10}^{-9}}{0.05\times {10}^{-3}}=0.01\text{rad}={0.57}^{\circ }$

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