If velocity of light in air is $$3$$×$$108$$ $$m/s$$ and that in water $$is2$$×$$108m/s$$ , then what would be the critical angle?

Correct option is 4sin−1(2/3)

Questions

If velocity of light in air is $3 \times 10^{8}$ m/s and that in water is $2 \times 10^{8} m / s$ , then what would be the critical angle?

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sin−1(3/2)

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detailed solution

Correct option is D

aμw=velocity of light in air velocity of light in water =3×1082×108=32 sinC=1aμw [where c is the critical angle ]∴ sinC=13/2=(23) or C=sin−1(23)

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If velocity of light in air is 3×108 m/s and that in water is2×108m/s , then what would be the critical angle?

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If velocity of light in air is $3 \times 10^{8}$ m/s and that in water is $2 \times 10^{8} m / s$ , then what would be the critical angle?