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 Question:
We have two lenses: a plano-convex lens of refractive index \( \mu_1 \) and a plano-concave lens of refractive index \( \mu_2 \). They fit exactly into each other, meaning their curved interface has the same radius of curvature \( R \). We need to find the equivalent focal length of the combination.

Step 2: Key Formula or Approach:
- Lens Maker's Formula:
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
- For a combination of two thin lenses in contact:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 3: Detailed Explanation:
Let's consider the two individual lenses:
1. Plano-convex lens (\( f_1 \)):
It has one flat surface (\( R_1 = \infty \)) and one convex surface of radius \( R_2 = -R \) (using Cartesian sign convention with light traveling from left to right):
\[ \frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = \frac{\mu_1 - 1}{R} \]
2. Plano-concave lens (\( f_2 \)):
Since it fits exactly onto the plano-convex lens, its left surface is concave with a radius of curvature \( R_1 = -R \) and its right surface is flat (\( R_2 = \infty \)):
\[ \frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) = -\frac{\mu_2 - 1}{R} \]
3. Focal length of the combination (\( F \)):
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
Substituting the values:
\[ \frac{1}{F} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} \]
\[ \frac{1}{F} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} \]
\[ \frac{1}{F} = \frac{\mu_1 - \mu_2}{R} \]
\[ F = \frac{R}{\mu_1 - \mu_2} \]

Step 4: Final Answer:
The focal length of the combination is \( F = \frac{R}{\mu_1 - \mu_2} \).