In a Young's double slit experiment, an angular width of the fringe is 0.35$^\circ$ on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is ${1 \over \alpha }$. The value of $\alpha$ is ___________.
Answer (integer)
4
Solution
<p>Angular fringe width $\theta = {\lambda \over D}$</p>
<p>So ${{{\theta _1}} \over {{\lambda _1}}} = {{{\theta _2}} \over {{\lambda _2}}}$</p>
<p>$${\theta _2} = {{0.35^\circ } \over {450\,nm}} \times {{450\,nm} \over {7/5}} = 0.25^\circ = {1 \over 4}$$</p>
<p>So $\alpha = 4$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
This question is part of PrepWiser's free JEE Main question bank. 197 more solved questions on Optics are available — start with the harder ones if your accuracy is >70%.