Question

If ${\mu }_0=\sqrt{2}$ and the ray just fail to emerge from the prism, refractive index $\mu$ of the solid will be

• A. $1.21$
• B. $\sqrt{2}$
• C. $1$
• D. $\sqrt{3}/2$

Answer 1

Answer: (C)

$1$

Using the relation obtained in the previous question, $\mu =\sqrt{{\mu }_o^2-si{n}^{2}i}$

As the ray just fails to emerge from the prism, so $i={90}^o$

Given : ${\mu }_o=\sqrt{2}$

$\mu =\sqrt{(\sqrt{2}{)}^2-si{n}^2{90}^o}$

Thus $\mu =1$