Question

A plano-convex lens fits exactly in to a plano-concave lens. Their plane surfaces are parallel to each other. Lenses are made up of different materials of refractive indices \( n_1 \) and \( n_2 \) and \( R \) is the radius of curvature of the curved surface of lenses. Focal length of the combination is

• A. \( \frac{R}{n_1 - n_2} \)
• B. \( \frac{R}{n_1 + n_2} \)
• C. \( \frac{R(n_1 - n_2)}{n_1n_2} \)
• D. \( \frac{R}{n_1 - n_2} \)

Hint:

For lenses in contact, the total focal length is the sum of the individual focal lengths. The focal length of a plano-convex lens depends on the refractive index and the radius of curvature.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have a combination of two lenses. We need to find the equivalent focal length of the system using Lens Maker's Formula for each component.
Step 2: Key Formula or Approach:
1. Lens Maker's Formula: \( \frac{1}{f} = (n - 1) (\frac{1}{R_1} - \frac{1}{R_2}) \).
2. Combination: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).
Step 3: Detailed Explanation:
For Plano-convex lens (\( n_1 \)):
Surfaces are \( R \) and \( \infty \).
\[ \frac{1}{f_1} = (n_1 - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{n_1 - 1}{R} \]
For Plano-concave lens (\( n_2 \)):
Surfaces are \( \infty \) and \( R \) (with appropriate sign).
\[ \frac{1}{f_2} = (n_2 - 1) \left( \frac{1}{\infty} - \frac{1}{R} \right) = -\frac{n_2 - 1}{R} \]
Equivalent focal length:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{n_1 - 1}{R} - \frac{n_2 - 1}{R} = \frac{n_1 - 1 - n_2 + 1}{R} \]
\[ \frac{1}{F} = \frac{n_1 - n_2}{R} \implies F = \frac{R}{n_1 - n_2} \]
Step 4: Final Answer:
The focal length is \( \frac{R}{n_1 - n_2} \).