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(A) $ {\sin ^{ - 1}}\left( {\dfrac{{10{t_2}}}{{{t_1}x}}} \right) $

(B) $ {\sin ^{ - 1}}\left( {\dfrac{{{t_2}x}}{{10{t_1}}}} \right) $

(C) $ {\sin ^{ - 1}}\left( {\dfrac{{10{t_1}}}{{{t_2}x}}} \right) $

(D) $ {\sin ^{ - 1}}\left( {\dfrac{{{t_1}x}}{{10{t_2}}}} \right) $

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At a particular angle of incidence called the critical angle, the refracted ray grazes the surface of separation. The angle of incidence in a denser medium for which the angle of refraction in a rarer medium is $ {90^0} $ is called the critical angle of the medium.

The time taken by the light to travel the distance $ x $ is $ {t_1} $ sec. Let $ c $ be the velocity of light through a vacuum.

$ c = \dfrac{x}{{{t_1}}} $

The time taken by the light to travel $ 10cm $ in a medium is taken as $ {t_2} $ . The velocity of light through the medium can be taken as $ v $

$ v = \dfrac{{10}}{{{t_2}}} $

The absolute refractive index can also be expressed as the ratio of the velocity of light in vacuum $ (c) $ to the velocity of light in the medium.

i.e.

$ \mu = \dfrac{c}{v} $

where $ \mu $ is the refractive index.

The critical angle is given by,

$ \sin C = \dfrac{1}{\mu } $

Substituting $ \mu = \dfrac{c}{v} $ in the above equation, we get

$ \sin C = \dfrac{1}{{\dfrac{c}{v}}} = \dfrac{v}{c} $

We know that,

$ c = \dfrac{x}{{{t_1}}} $

And

$ v = \dfrac{{10}}{{{t_2}}} $

Substituting these values we get,

$ \sin C = \dfrac{{\dfrac{{10}}{{{t_2}}}}}{{\dfrac{x}{{{t_1}}}}} = \dfrac{{10{t_1}}}{{x{t_2}}} $

From this, we can write the critical angle as,

$ C = {\sin ^{ - 1}}\left( {\dfrac{{10{t_1}}}{{{t_2}x}}} \right) $

The velocity of light through any medium will be less than that of the velocity of light through a vacuum. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and a given colour of light. This law is called Snell’s law.