JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 21)
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
4
Explanation
Angular fringe width $$\theta = {\lambda \over D}$$
So $${{{\theta _1}} \over {{\lambda _1}}} = {{{\theta _2}} \over {{\lambda _2}}}$$
$${\theta _2} = {{0.35^\circ } \over {450\,nm}} \times {{450\,nm} \over {7/5}} = 0.25^\circ = {1 \over 4}$$
So $$\alpha = 4$$
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