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Question

Physics Question on Ray optics and optical instruments

If the critical angle for total internal reflection from a medium to vacuum is 3030{}^\circ , then the speed of light in the medium is

A

6×108m/s6\times 10^{8}m/s

B

3×108m/s3\times 10^{8}\, m/s

C

2×108m/s2\times 10^{8}\,m/s

D

1.5×108m/s1.5\times 10^{8}m/s

Answer

1.5×108m/s1.5\times 10^{8}m/s

Explanation

Solution

Critical angle C=30C=30^{\circ} Refractive indexroman μ=1sinC=1sin30=2\mu=\frac{1}{\sin C}=\frac{1}{\sin 30^{\circ}}=2 \therefore Also μ=cv \mu =\frac{c}{v} \therefore speed of light in medium v=cμ=3×1082v =\frac{c}{\mu}=\frac{3 \times 10^{8}}{2} =1.5×108m/s=1.5 \times 10^{8} m / s