Q.
The angle of minimum deviation for a thin prism with respect to air and when dipped in water will be:
$\left({ }_{a} \mu_{g}=\frac{3}{2}, {}_{a} \mu_{w}=\frac{4}{3}\right)$
Jharkhand CECEJharkhand CECE 2003
Solution:
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
$\frac{\delta_{m}}{\delta_{m}}=\frac{1}{4}$
