Material $A$has critical angle $i_{A},$ and material $B$ has critical angle $i_{B}\left( i_{b} i_{A} \right).$ Then which of the following is true? (i) Light can be totally internally reflected when it passes from $B\ to\ A$ (ii) Light can be totally internally reflected when it passes from $A\ to\ B$ (iii) Critical angle for total internal reflection is $i_{B} - i_{A}$ (iv) Critical angle between $A\ and\ B$ is $\sin^{- 1}\left( \frac{\sin i_{A}}{\sin i_{B}} \right)$

Question

Material $A$has critical angle $i_{A},$ and material $B$ has critical angle $i_{B}\left( i_{b} > i_{A} \right).$ Then which of the following is true? (i) Light can be totally internally reflected when it passes from $B\ to\ A$ (ii) Light can be totally internally reflected when it passes from $A\ to\ B$ (iii) Critical angle for total internal reflection is $i_{B} - i_{A}$ (iv) Critical angle between $A\ and\ B$ is $\sin^{- 1}\left( \frac{\sin i_{A}}{\sin i_{B}} \right)$

Dr. Abubacker · Accepted Answer

(d) From Snell’s law
$\mu = \frac{1}{\sin i_{C}}$
Since, $i_{B}\  > i_{A}$
$\sin{i_{:B} > \sin i_{A}}$
$\Rightarrow \mu_{B} < \mu_{A}$
When angle of incidence in the denser medium is increased even very slightly beyond the critical angle, then the ray of light is reflected back completely in the denser medium and total internal reflection takes place. Let $	heta$ be critical angle from medium $A\ to\ B,\ $then
$\ _{B}\mu_{A} = \frac{1}{\sin	heta}$
$\Rightarrow \sin	heta = \frac{1}{\ _{B}\mu_{A}}\frac{\mu_{B}}{\mu_{A}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \frac{1}{\sin i_{B}} 	imes \frac{\sin i_{A}}{1}$
$\Rightarrow \ 	heta = \sin^{- 1}\left\lbrack \frac{\sin i_{A}}{\sin i_{B}} ightbrack$

Explanation

Reviewed by Dr. Abubacker

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