Diameter of a plano–convex lens is 6 cm and thickness at the centre is 3 mm. If the speed of light in the material of the lens is 2 $\times$10⁸ m/sec, the focal length of the lens is

According to lens formula

         $\frac{1}{f}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$

         The lens is plano-convex i.e.,

         $R_1=R$ and $R_2=\infty$

         Hence $\frac{1}{f}=\frac{\mu -1}{R}\Rightarrow f=\frac{R}{\mu -1}$

         Speed of light in medium of lens

         $v=2\times {10}^{8}m/s$;     $\mu =\frac{c}{v}=\frac{3\times {10}^8}{2\times {10}^8}=\frac{3}{2}=1.5$

If r is the radius and y is the thickness of lens (at the centre), the radius of curvature R of its curved surface in accordance with the figure is given by

$R^2=r^2+{(R-y)}^2\Rightarrow r^2+y^2-2Ry=0$

Neglecting $y^2$;  

we get $R=\frac{r^2}{2y}=\frac{{(6/2)}^2}{2\times 0.3}=15cm$ 

Hence $f=\frac{15}{1.5-1}=30cm$