A) \[{{\sin }^{-1}}\left( \frac{10{{t}_{1}}}{{{t}_{2}}} \right)\]
B) \[{{\sin }^{-1}}\left( \frac{{{t}_{1}}}{10{{t}_{2}}} \right)\]
C) \[{{\sin }^{-1}}\left( \frac{{{t}_{2}}}{10{{t}_{1}}} \right)\]
D) \[{{\sin }^{-1}}\left( \frac{10{{t}_{2}}}{{{t}_{1}}} \right)\]
Correct Answer: A
Solution :
Here, Initial time\[={{t}_{1}}\]sec Initial distance\[{{x}_{1}}=x\,cm\] Final time\[={{t}_{2}}\]sec Final distance\[{{x}_{2}}=10x\,cm\] The velocity of light in vacuum is \[{{\upsilon }_{0}}=\frac{x}{{{t}_{1}}}\] The velocity of light in medium is \[\upsilon =\frac{10x}{{{t}_{2}}}\] Now, the relation for critical angle is given by, \[\sin C=\frac{1}{\mu }\] So, \[C={{\sin }^{-1}}\frac{1}{\mu }\] and refractive index\[\mu \]is given by, \[=\frac{{{\upsilon }_{0}}}{\upsilon }=\frac{x}{{{t}_{1}}}\times \frac{{{t}_{2}}}{10x}=\frac{{{t}_{2}}}{10{{t}_{1}}}\] Hence, \[C={{\sin }^{-1}}\frac{1}{\mu }={{\sin }^{-1}}\left( \frac{10{{t}_{1}}}{{{t}_{2}}} \right)\]
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