A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. I

Question

A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices $\mu_1$ and $\mu_2$ and $R$ is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is:

TopperSquare Expert · Accepted Answer

The correct answer is Option (C): $\frac{R}{\mu_1-\mu_2}$. 1.  **Focal Length of Plano-Convex Lens ($f_1$):**
    Let's assume light enters the plano-convex lens through its plane surface. The radii of curvature are $R_1 = \infty$ and $R_2 = -R$ (using sign convention). Using the Lens Maker's formula:
    $\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{\mu_1 - 1}{R}$.

2.  **Focal Length of Plano-Concave Lens ($f_2$):**
    The plano-concave lens fits exactly over the curved surface, so its curved surface has the same radius $R$ but is concave towards the incident light. The plane surfaces are parallel. Its radii are $R_1 = -R$ and $R_2 = \infty$.
    $\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{\mu_2 - 1}{R}$.

3.  **Equivalent Focal Length ($F$):**
    Since the lenses are in contact, the equivalent focal length of the combination is obtained by adding their optical powers:
    $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
    $\frac{1}{F} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{(\mu_1 - 1) - (\mu_2 - 1)}{R} = \frac{\mu_1 - \mu_2}{R}$.
    
    Therefore, the focal length of the combination is $F = \frac{R}{\mu_1 - \mu_2}$.

Step-by-Step Solution

Exam Context & Concepts Covered

Everything you need to crack NEET

Track Your Progress

Platform Features

15,000+ MCQs

AI Analytics

Detailed Solutions