In a Youngs double slit | Wave Optics

JEE MAIN 2022

28-06-22 S2

Question

In a Youngs 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 . The value of $\alpha$ is $\_\_\_\_$

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Solution

$\begin{aligned} & \beta=\frac{0.35 \times 5}{7}=0.25 \\ & \frac{1}{\alpha}=\frac{25}{100} \\ & \alpha=4\end{aligned}$

Question Tags

JEE Main

Physics

Easy

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