Question 101 of 295

Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt{2n}$. The incident angle is twice that of refracting angle. Then, the angle of incidence will be

A${\sin }^{-1}\sqrt{n}$B${\cos }^{-1}\left(\dfrac{n}{\sqrt{2}}\right)$C${\sin }^{-1}\left(\sqrt{2n}\right)$D$2{\cos }^{-1}\left(\sqrt{\dfrac{n}{2}}\right)$

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Answer: (D) $2{\cos }^{-1}\left(\sqrt{\dfrac{n}{2}}\right)$

Let angle of incidence be $\theta$, then angle of refraction $\dfrac{\theta }{2}$. From Snell's law, we can write,
$1\times \sin \theta =\sqrt{2n}\times \sin \dfrac{\theta }{2}$, $2\sin \left(\dfrac{\theta }{2}\right)\cos \left(\dfrac{\theta }{2}\right)=\sqrt{2n}\sin \left(\dfrac{\theta }{2}\right)$
$\cos \left(\dfrac{\theta }{2}\right)=\sqrt{\dfrac{n}{2}}\Rightarrow \theta =2{\cos }^{-1}\left(\sqrt{\dfrac{n}{2}}\right)$