Question

In the figure, a light ray is incident on the lower medium boundary at an angle of $45$ the normal. Which of the following statements are true?

• A. If ${\mu }_2>\sqrt{2},$ then angle of deviation is ${45}^{\circ }$
• B. If ${\mu }_2<\sqrt{2},$ then angle of deviation is ${90}^{\circ }$
• C. If ${\mu }_2<\sqrt{2},$ then angle of deviation is ${135}^{\circ }$
• D. If ${\mu }_2>\sqrt{2}$ , then angle of deviation is $0^0$

Answer 1

Answer: (A), (B)

The correct options are
A If ${\mu }_2>\sqrt{2},$ then angle of deviation is ${45}^{\circ }$
B If ${\mu }_2<\sqrt{2},$ then angle of deviation is ${90}^{\circ }$

There will be two conditions possible for the given situation

For the first condition where total internal reflection will not take place,

${\mu }_2>\sqrt{2}$

From the figure,

Applying Snell’s law

${\mu }_1\sin i={\mu }_2\sin r$

Hence,

$2\sin {45}^{\circ }={\mu }_2\sin r$ …… (1)

Similarly,

${\mu }_2\sin r={\mu }_3\sin e$

$\therefore {\mu }_2\sin r=\sqrt{2}\sin e$ …… (2)

From the equation (1) and (20

$2\sin {45}^{\circ }=\sqrt{2}\sin e$

$\sin e=1\Rightarrow e={90}^{\circ }$

Therefore the deviation will be ${45}^{\circ }$

Now for the second condition that the total internal reflection will not take place

${\mu }_2<\sqrt{2}$

Hence for this condition, it can be seen from the figure

The deviation will be ${90}^{\circ }$

The correct answer is option (A) and (B).