Question

if speed of light in water is 2.2*10^8 , then the critical angle of incidence for light going from water to air is

Answer 1

Answer:

Speed of light in vacuum = 3.0 x 108m/s

Speed of light in water = 2.25 x 108m/s

Refractive index of water = ?

We know that

Refractive index of water = speed of light in vacuum/speed of light in water

Refractive index of water = 3 x 108/2.25 x 108 = 1.33

Explanation:

Kashuuu~

Answer 2

The critical angle of incidence for light is 41.3.

Given - Speed of light

Find - Critical angle of incidence

Solution - Firstly finding refractive index followed by calculating critical angle of incidence.

Refractive index = speed of light in air/speed of light in water.

Refractive index = $\frac{3 \times {10}^{8} }{2.2 \times {10}^{8} }$

Refractive index = 1.5

Critical angle = ${ \sin }^{ - 1}$ 1/ refractive index

Critical angle = ${ \sin}^{ - 1}$ $\frac{1}{1.5}$

Critical angle = 41.3

Hence, critical angle of incidence is 41.3.

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