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 \(2  \times  10^8 m/sec\), the focal length of the lens is.......\(cm\)

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

Hence \(f = \frac{{15}}{{1.5 - 1}} = 30 cm\)