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Question

Physics Question on Ray optics and optical instruments

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

A

3×108m/s3\times {{10}^{8}}\,m/s

B

1.5×108m/s1.5\times {{10}^{8}}\,m/s

C

6×108m/s6\times {{10}^{8}}\,m/s

D

3×108m/s\sqrt{3}\,\times {{10}^{8}}\,m/s

Answer

1.5×108m/s1.5\times {{10}^{8}}\,m/s

Explanation

Solution

Refractive index medium μ=1sinC\mu=\frac{1}{\sin C} where CC is critical angle. Given, C=30C=30^{\circ} μ=1sin30=11/2=2 \therefore \mu=\frac{1}{\sin 30^{\circ}}=\frac{1}{1 / 2}=2 From Snell's law μ=v0vm\mu=\frac{v_{0}}{v_{m}} where v0v_{0} is speed of light in vacuum and vmv_{m} the velocity in medium. vm=3×1082\therefore v_{m}=\frac{3 \times 10^{8}}{2} =1.5×108m/s=1.5 \times 10^{8}\, m / s