$(a)\; n > \frac{1}{\sqrt 2} \\ (b)\;n < \frac{1}{\sqrt 2} \\ (c)\; n > \sqrt 2 \\ (d)\; n < \sqrt 2 $

It is required that no ray exits from the curved surface.

This will be automatically fulfilled if minimum r' is more than critical angle.

Angle r' is minimum when r is maximum ie C.

Therefore minimum value of r' is $90-c$

From above condition

$90 ^{\circ} -C > C$

or $C < 45^{\circ}$

$\sin C < \sin 45 ^{\circ}$

$\large\frac{1}{n} < \frac{1}{\sqrt 2}$

$n > \sqrt 2$

Hence c is the correct answer.

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