Ray Optics and Optical Instruments Part 1

Given, thickness at the centre of plano-convex lens, $t = 3\ \mathrm{mm}$ $= 3 \times 10^{-3}\ \mathrm{m}$ Diameter of plano-convex lens, $d = 6\ \mathrm{cm}$ $\therefore$ Radius of plano-convex lens, $r = 3\ \mathrm{cm} = 3 \times 10^{-2}\ \mathrm{m}$ Speed of light in lens material, $v = 2 \times 10^{8}\ \mathrm{ms}^{-1}$ Refractive index $= \frac{\text{Speed of light in air}}{\text{Speed of light in medium}}$ $\mu = \frac{3 \times 10^{8}}{2 \times 10^{8}} = \frac{3}{2} = 1.5$ $\Rightarrow \mu = 1.5$We know that, $R^{2} = (R - t)^{2} + r^{2}$ where, $R$ is the radius of curvature of plano convex lens. As, $t$ is small, then $t^{2}$ will be very very small, so it can be neglected. $\Rightarrow 2Rt = r^{2}$ $\Rightarrow R = \frac{r^{2}}{2t}$....(i) From lens Maker's formula, we have $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$ Here $R_{1} = R \text{ and } R_{2} = \infty$ $\Rightarrow \frac{1}{f} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right)$ $\Rightarrow \frac{1}{f} = \frac{\mu - 1}{R}$...(ii) From Eqs. (i) and (ii), we get $\frac{1}{f} = \frac{\mu - 1(2t)}{r^{2}} \Rightarrow f = \frac{r^{2}}{(\mu - 1)2t}$ Putting the given values in above equation, we get $f = \frac{(3 \times 10^{-2})^{2}}{(1.5 - 1)2 \times 3 \times 10^{-3}} \Rightarrow f = 0.3\ \mathrm{m}$ $f = 30\ \mathrm{cm}$