Question

Angle of minimum deviation produced by a thin prism $(\mu =3/2)$ is $4^{\circ }$. If the prism is immersed in water $({\mu }_w=4/3)$, the angle of minimum deviation will be

• A. $4^{\circ }$
• B. ${16}^{\circ }$
• C. $8^{\circ }$
• D. $1^{\circ }$

Answer 1

The correct option is D $1^{\circ }$

The angle of minimum deviation is given by the formula,

${\delta }_a=\left(\frac{{\mu }_g}{{\mu }_a}-1\right)A$

Substituting the given values of refractive indices ${\mu }_a=1$ for air, and ${\mu }_g=3/2$, hence

${\delta }_a=\left(\frac{3}{2}-1\right)A\Rightarrow {\delta }_a=\frac{1}{2}A$

When immersed in water, the angle of minimum deviation is given as

${\delta }_w=\left(\frac{{\mu }_g}{{\mu }_w}-1\right)A$

Substituting the given values of refractive indices ${\mu }_a=4/3$ for air, and ${\mu }_g=3/2$, hence

${\delta }_w=\left(\frac{3/2}{4/3}-1\right)A\Rightarrow {\delta }_w=\left(\frac{9}{8}-1\right)A\Rightarrow {\delta }_w=\frac{1}{8}A$

It can be concluded that the ${\delta }_w$ is $\frac{1}{4}th$ times than the ${\delta }_a$, hence

${\delta }_w=\frac{1}{4}{\delta }_a$

Given that ${\delta }_a=4^{\circ }$

${\delta }_w=\frac{1}{4}\times 4^{\circ }$

${\delta }_w=1^{\circ }$

The angle of minimum deviation when immersed in water is $1^{\circ }$.