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:

For the first planoconvex lens:

$$\mu_1 = 1.5, \quad R_1 = 15\text{ cm}, \quad R_2 = \infty$$

$$P_1 = \frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \implies P_1 = (1.5 - 1)\left(\frac{1}{15} - \frac{1}{\infty}\right) = \frac{1}{30}\text{ cm}^{-1}$$

For the second planoconvex lens:

$$\mu_2 = 1.2, \quad R_3 = \infty, \quad R_4 = -12\text{ cm}$$

$$P_2 = \frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{R_3} - \frac{1}{R_4}\right) \implies P_2 = (1.2 - 1)\left(\frac{1}{\infty} - \frac{1}{-12}\right) = \frac{1}{60}\text{ cm}^{-1}$$

Equivalent focal length of the combination:

$$P_{\text{eq}} = P_1 + P_2 \implies \frac{1}{f_{\text{eq}}} = \frac{1}{30} + \frac{1}{60} = \frac{1}{20}\text{ cm}^{-1} \implies f_{\text{eq}} = 20\text{ cm}$$

Using thin lens formula:

$$u = -30\text{ cm}$$

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{\text{eq}}} \implies \frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \implies v = 60\text{ cm}$$

Magnification: $$m = \frac{v}{u} \implies m = \frac{60}{-30} = -2$$