Ray Optics and Optical Instruments Part 3

Concept:A biconvex lens can be considered as two thin planoconvex lenses in contact. Their combined focal length is found by adding their powers.Explanation:For left planoconvex lens $\mathrm{L}_1$: refractive index $\mu_1 = 1.5$, curved surface radius $R_1 = +15\ \text{cm}$ (convex towards left), plane surface $R_2 = \infty$.Using Lens Maker’s formula: $\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (0.5)\left(\frac{1}{15} - \frac{1}{\infty}\right) = \frac{0.5}{15} = \frac{1}{30}$.Thus $f_1 = +30\ \text{cm}$.For right planoconvex lens $\mathrm{L}_2$: refractive index $\mu_2 = 1.2$, plane surface $R_1 = \infty$, curved surface radius $R_2 = -12\ \text{cm}$ (convex towards right, so sign negative as per sign convention).Using Lens Maker’s formula: $\frac{1}{f_2} = (1.2 - 1)\left(\frac{1}{\infty} - \frac{1}{-12}\right) = (0.2)\left(0 + \frac{1}{12}\right) = \frac{0.2}{12} = \frac{1}{60}$.Thus $f_2 = +60\ \text{cm}$.Since the lenses are thin and in contact, the combined focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.$\frac{1}{F} = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20}$.So $F = +20\ \text{cm}$.Object distance $u = -30\ \text{cm}$ (sign convention).Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$.$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$$\frac{1}{v} + \frac{1}{30} = \frac{1}{20}$$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$Therefore $v = +60\ \text{cm}$.Magnification $m = \frac{v}{u} = \frac{60}{-30} = -2$.The negative sign means the image is real and inverted, and twice the size of the object.Answer:A. $-2$