Calculate the radius of curvature of an equiconcave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of -5D?
OR 
An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index $4\sqrt{2}/5$
 

In an equiconcave lens, radius of curvature of both surfaces are equal
$\begin{array}{l}\therefore \frac{1}{\mathrm{f}}=(\mathrm{\mu }-1)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right)\\ \text{ since, }\mathrm{P}=-5\mathrm{D}\\ \therefore \frac{1}{\mathrm{f}\left(\mathrm{m}\right)}=\mathrm{P}=-5\\ \text{ and }{\mathrm{\mu }}_{\mathrm{e}}=1.5\text{ and }{\mathrm{\mu }}_{\mathrm{m}}=1.4\\ \therefore 5=\left(\frac{1.5}{1.4}-1\right)\left(\frac{-2}{\mathrm{R}}\right)\\ \Rightarrow -5=\frac{0.1}{1.4}\times \frac{-2}{\mathrm{R}}\\ \Rightarrow \mathrm{R}=\frac{0.1\times (-2)}{1.4\times (-5)}\\ =0.03\mathrm{m}\\ \end{array}$
OR 
As we know,
${ }_1{\mathrm{n}}_2=\frac{\sin \left(\frac{\mathrm{A}+{\mathrm{\delta }}_{min}}{2}\right)}{\sin \frac{\mathrm{A}}{2}}$
where n₂=1.6, refractive index of glass prism
${\mathrm{n}}_1=\frac{4\sqrt{2}}{5}$, refractive index of medium
$\mathrm{A}={60}^{\circ }, \mathrm{angle} \mathrm{of} \mathrm{prism}$
$\begin{array}{l}\therefore \frac{1.6}{4\sqrt{2}/5}=\sin \frac{\left(\frac{{60}^{\circ }+{\mathrm{\delta }}_{\mathrm{m}}}{2}\right)}{\sin 30}\\ \Rightarrow \frac{0.4\times 5}{\sqrt{2}}=\sin \left(\frac{{60}^{\circ }+{\mathrm{\delta }}_{\mathrm{m}}}{2}\right)\times 2\left[\because \sin {30}^{\circ }=\frac{1}{2}\right]\\ \Rightarrow \sin \left(\frac{{60}^{\circ }+{\mathrm{\delta }}_{\mathrm{m}}}{2}\right)=\frac{1}{\sqrt{2}}\\ \Rightarrow \frac{{60}^{\circ }+{\mathrm{\delta }}_{\mathrm{m}}}{2}={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)\\ ={\sin }^{-1}\left(\sin {45}^{\circ }\right)={45}^{\circ }\\ \Rightarrow {\mathrm{\delta }}_{\mathrm{m}}={30}^{\circ }\\ \end{array}$