The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a piano-concave lens (refractive index = l.7) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back smface of the biconvex lens is________.

For biconvex lens:

$$\mu_1 = 1.5$$

$$P_1 = (\mu_1 - 1)\left(\frac{1}{R_1} - \frac{1}{-R_2}\right)$$

$$P_1 = (1.5 - 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

$$P_1 = 0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

For plano-concave lens:

$$\mu_2 = 1.7$$

$$R_3 = -R_2$$

$$R_4 = \infty$$

$$P_2 = (\mu_2 - 1)\left(\frac{1}{-R_2} - \frac{1}{\infty}\right)$$

$$P_2 = (1.7 - 1)\left(-\frac{1}{R_2}\right)$$

$$P_2 = -\frac{0.7}{R_2}$$

Given:

$$|P_1| = |P_2|$$

$$0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2}$$

$$0.5\frac{1}{R_1} + 0.5\frac{1}{R_2} = 0.7\frac{1}{R_2}$$

$$0.5\frac{1}{R_1} = 0.2\frac{1}{R_2}$$

$$\frac{R_1}{R_2} = \frac{0.5}{0.2}$$

$$\frac{R_1}{R_2} = \frac{5}{2}$$