Question

The angle of minimum deviation for a thin prism with respect to air and when dipped in water will be:
$\left({}_a{\mu }_g=\frac{3}{2},{}_a{\mu }_w=\frac{4}{3}\right)$

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

For a prism of refractive index $\mu$,
angle of minimum deviation is given by
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$
When prism is thin, ${\delta }_m$ is small and
$\frac{\sin A+{\delta }_m}{2}=\frac{A+{\delta }_m}{2}$
and $\sin \frac{A}{2}=\frac{A}{2}$
$\therefore \mu =\frac{\left(A+{\delta }_m\right)/2}{A/2}$
$\Rightarrow {\delta }_m-(\mu -1)A$
When in air, ${\delta }_m=\left({}_a{\mu }_g-1\right)A$
${\delta }_m=\left(\frac{3}{2}-1\right)A=\frac{A}{2}..$ (i)
When dipped in water
${\delta }_m=\left({}_w{\mu }_g-1\right)A$
$=\left(\frac{a{\mu }_g}{a{\mu }_w}-1\right)A$
${\delta }_m=\left(\frac{3/2}{4/3}-1\right)A=\frac{A}{8}..$ (ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{{\delta }_m}{{\delta }_m}=\frac{1}{4}$