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

i=2 r sini×n1=sinr×n2 sini×1=sini2×2n sinisini2=2n 2sini2cosi2sini2=2n cosi2=n2 i2=cos-1n2 i=2cos-1n2

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

i=2 r sini×n1=sinr×n2 sini×1=sini2×2n sinisini2=2n 2sini2cosi2sini2=2n cosi2=n2 i2=cos-1n2 i=2cos-1n2

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

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$\sin^{- 1} (\sqrt{n})$

$2 \cos^{- 1} (\sqrt{\frac{n}{2}})$

$\cos^{- 1} (\sqrt{\frac{n}{2}})$

$\sin^{- 1} (\sqrt{2 n})$

answer is D.

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Detailed Solution

$i = 2 r \sin i \times n_{1} = \sin r \times n_{2} \sin i \times 1 = \sin \frac{i}{2} \times \sqrt{2 n} \frac{\sin i}{\sin \frac{i}{2}} = \sqrt{2 n} \frac{2 \sin \frac{i}{2} \cos \frac{i}{2}}{\sin \frac{i}{2}} = \sqrt{2 n} \cos \frac{i}{2} = \sqrt{\frac{n}{2}} \frac{i}{2} = \cos^{- 1} (\sqrt{\frac{n}{2}}) i = 2 \cos^{- 1} (\sqrt{\frac{n}{2}})$