CBSE 2025, Compartment, Set 1

Question

A plano-convex lens of refractive index $\mu_1$ is placed coaxially in contact with a biconcave lens of refractive index $\mu_2$ as shown in the figure. All curved faces are of radius of curvature $R$ each. Obtain the expression for the focal length of the combined lens.

![Plano-convex and biconcave lens](https://www.edvaya.com/images/cbse/physics/2025/plano-convex-biconcave-lens.webp)

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** By the Lens Maker's formula, $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. **For the Plano-convex lens ($f_1$):** The first surface is plane ($R_1 = \infty$), and the second surface is curved inward ($R_2 = -R$). $$\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right)$$ $$\frac{1}{f_1} = \frac{(\mu_1 - 1)}{R}$$ **For the Biconcave lens ($f_2$):** The first surface is curved inward ($R_1 = -R$), and the second surface is curved outward ($R_2 = +R$). $$\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-R} - \frac{1}{R}\right)$$ $$\frac{1}{f_2} = -2\frac{(\mu_2 - 1)}{R}$$ **Equivalent focal length of combination ($f$):** $$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$$ $$\frac{1}{f} = \frac{\mu_1 - 1}{R} - \frac{2(\mu_2 - 1)}{R}$$ $$\frac{1}{f} = \frac{\mu_1 - 1 - 2\mu_2 + 2}{R}$$ $$\frac{1}{f} = \frac{\mu_1 - 2\mu_2 + 1}{R}$$ Therefore, the focal length is: $$f = \frac{R}{\mu_1 - 2\mu_2 + 1}$$

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