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

• A. \( \frac{R}{\mu_1 - \mu_2} \)
• B. \( \frac{2R}{\mu_1 - \mu_2} \)
• C. \( \frac{R}{2(\mu_1 - \mu_2)} \)
• D. \( \frac{R}{\mu_1 + \mu_2} \)

Hint:

This combination can be treated as a single lens with a curved interface separating two different media, or simply as two thin lenses in contact. Using the Lens Maker's Formula with careful sign conventions for each lens is the most reliable way to avoid sign errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When two thin lenses are placed in contact, the effective power of the combination is the algebraic sum of the powers of the individual lenses.
The power \(P\) of a lens is defined as the reciprocal of its focal length \(f\), i.e., \(P = 1/f\).
For the combination, the total power is \(P_{eq} = P_1 + P_2\).
The individual powers are determined using the Lens Maker's Formula.
Since the lenses "fit exactly" and the final setup has parallel plane surfaces, the curved surfaces of both lenses must share the same radius of curvature \(R\).
Step 2: Key Formula or Approach:
Lens Maker's Formula: \(\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\).
Sign Convention: Light travels from left to right. Surfaces that are convex towards the light have positive radii, and surfaces concave towards the light have negative radii.
Combination formula: \(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\).
Step 3: Detailed Explanation:
Consider the first lens to be the plano-convex lens with refractive index \(\mu_1\).
Let its plane surface be the first surface (\(R_{1a} = \infty\)) and the curved surface be the second surface (\(R_{1b} = -R\)).
Using Lens Maker's formula for the first lens:
\[ \frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{\mu_1 - 1}{R} \]
Now, consider the second lens to be the plano-concave lens with refractive index \(\mu_2\).
For it to fit exactly and have its plane surface parallel to the first, its curved surface must be the first surface (\(R_{2a} = -R\)) and its plane surface must be the second surface (\(R_{2b} = \infty\)).
Using Lens Maker's formula for the second lens:
\[ \frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{\mu_2 - 1}{R} \]
Now, find the reciprocal of the equivalent focal length \(F\):
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
\[ \frac{1}{F} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} \]
Combine the terms over the common denominator \(R\):
\[ \frac{1}{F} = \frac{(\mu_1 - 1) - (\mu_2 - 1)}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} \]
\[ \frac{1}{F} = \frac{\mu_1 - \mu_2}{R} \]
Taking the reciprocal to find \(F\):
\[ F = \frac{R}{\mu_1 - \mu_2} \]
This matches the expression in option (A).
Step 4: Final Answer:
The equivalent focal length depends on the difference between the refractive indices of the two materials.
The final result is \(F = \frac{R}{\mu_1 - \mu_2}\).