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Question: Light takes \('{t_1}'\) sec to travel a distance \('x'\) in vacuum and the same light takes \('{t_2}...

Light takes ′t1′'{t_1}' sec to travel a distance ′x′'x' in vacuum and the same light takes ′t2′'{t_2}' sec to travel 10cm10cm in a medium. Critical angle for corresponding medium will be:
A) sin⁡−1(10t2t1x){\sin ^{ - 1}}\left( {\dfrac{{10{t_2}}}{{{t_1}x}}} \right)
B) sin⁡−1(t2x10t1){\sin ^{ - 1}}\left( {\dfrac{{{t_2}x}}{{10{t_1}}}} \right)
C) sin⁡−1(10t1t2x){\sin ^{ - 1}}\left( {\dfrac{{10{t_1}}}{{{t_2}x}}} \right)
D) sin⁡−1(t1x10t2){\sin ^{ - 1}}\left( {\dfrac{{{t_1}x}}{{10{t_2}}}} \right)

Explanation

Solution

Critical angle refers to an angle of incidence at which total internal reflection occurs. Light travels with different speeds in different mediums and the ratio of these speeds is known as refractive index.

Complete step by step answer:
Refractive index is the ratio of speed of light first medium to second medium. These media can vary indifferently. Relative refractive index is the ratio of speed of light in two media. Whereas the absolute refractive index of any medium is the ratio of speed of light in vacuum to speed of light in that medium.
In this case, let the speed of light through vacuum be cc. Then,
⇒c=xt1\Rightarrow c = \dfrac{x}{{{t_1}}}
And, let the speed to light through the medium be vv. Then,
⇒v=10t2\Rightarrow v = \dfrac{{10}}{{{t_2}}}
So absolute refractive index(μ)\left( \mu \right) of the medium can be calculated by,
⇒μ=cv\Rightarrow \mu = \dfrac{c}{v}
⇒μ=xt110t2\Rightarrow \mu = \dfrac{\dfrac{x}{t_1}}{\dfrac{10}{t_2}}
⇒μ=xt1×t210\Rightarrow \mu = \dfrac{x}{{{t_1}}} \times \dfrac{{{t_2}}}{{10}}
⇒μ=t2x10t1\Rightarrow \mu = \dfrac{{{t_2}x}}{{10{t_1}}}
Critical angle is the angle of incidence at which a rays travelling from one medium to another starts undergoing total internal reflection. Critical Angle(θC)\left( {{\theta _C}} \right) is given by, sin⁡θC=1μ\sin {\theta _C} = \dfrac{1}{\mu }
⇒sin⁡θC=1t2x10t1\Rightarrow \sin {\theta _C} = \dfrac{1}{\dfrac{t_2 x}{10 t_1}}
⇒sin⁡θC=10t1t2x\Rightarrow \sin {\theta _C} = \dfrac{{10{t_1}}}{{{t_2}x}}
⇒θc=sin⁡−1(10t1t2x)\Rightarrow {\theta _c} = {\sin ^{ - 1}}\left( {\dfrac{{10{t_1}}}{{{t_2}x}}} \right)

Hence, the correct answer is option C.

Note: Total Internal Reflection has a lot of implications in our daily life like optical fibres, formation of a rainbow, working of a prism, and the list goes on. In optical fibres, when the light ray strikes the fibre at a critical angle, it goes on reflecting indefinitely, and that’s how information travels really long distances within seconds of time.