Figure, a light ray is incident on the lower medium boundary at an angle of 45^o with the normal. Which of the following statements is/are true?

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If $μ_{2} < \sqrt{2}$ , then angle of deviation is 135^o

If $μ_{2} > \sqrt{2}$ , then angle of deviation is 45^o

If $μ_{2} > \sqrt{2}$ , then angle of deviation is 0^o

If $μ_{2} < \sqrt{2}$ , then angle of deviation is 90^o

answer is A, B.

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Detailed Solution

If $μ_{2} = \sqrt{2}, c r i t i c a l a n g l e = 45^{0}$

For $μ_{2} > \sqrt{2}, c r i t i c a l a n g l e > 45^{}$

Hence for $μ_{2} > \sqrt{2}$ TIR will not take Place

$\begin{array}{l} 2 \sin 45^{\circ} = μ_{2} \sin r \\ μ_{2} \sin r = \sqrt{2} \sin e \\ e = 90^{\circ} \end{array}$

Hence, deviation = e - $45^{0} =$ 45^o

For $μ_{2} < \sqrt{2}$ , critical angle = $45^{0} a n d$ TIR will take place.

Deviation is 90^o.

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