A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be :
Answer: (A) $-2$
Two thin lenses in contact: $\dfrac1f = \dfrac1{f_1}+\dfrac1{f_2}$.
Left half ($\mu = 1.5$, $R = 15$ cm): $\dfrac1{f_1} = (1.5-1)\dfrac1{15} = \dfrac1{30}$
Right half ($\mu = 1.2$, $R = 12$ cm): $\dfrac1{f_2} = (1.2-1)\dfrac1{12} = \dfrac1{60}$
$$\frac1f = \frac1{30}+\frac1{60} = \frac1{20}\Rightarrow f = 20\ \text{cm}$$
With $u = -30$ cm: $\dfrac1v = \dfrac1f + \dfrac1u = \dfrac1{20}-\dfrac1{30} = \dfrac1{60}$, so $v = 60$ cm.
$$m = \frac vu = \frac{60}{-30} = -2$$
Solution by Sreeraj P, M.Sc Physics