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Light takes $ {t_1} $ sec to travel a distance $ x $ in a vacuum and the same light takes $ {t_2} $ sec to travel $ 10cm $ in a medium. The critical angle for the corresponding medium will be
(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) $

Answer
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Hint: When light passes from one medium to another it will bend its path. This phenomenon is called refraction. The path of the ray undergoing reflection or refraction is reversible. This is called the principle of reversibility of light. The refractive index of the medium concerning air or vacuum is called the absolute refractive index.

Complete step by step solution
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 answer is: Option (C); $ {\sin ^{ - 1}}\left( {\dfrac{{10{t_1}}}{{{t_2}x}}} \right) $ .

Note
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.