Question

What is the focal length of double concave lens kept in air with two spherical surfaces radii R₁= 20 cm and R₂ = 40 cm. Take refractive index of lens n = \(\frac{5}{3}\)

• A. -20 cm
• B. 20 cm
• C. 40 cm
• D. -40 cm

Answer 1

The Correct Option is A

To solve the problem, we need to calculate the focal length of a double concave lens placed in air using the lens maker’s formula. The radii of curvature are given as \( R_1 = 20 \, \text{cm} \) and \( R_2 = 40 \, \text{cm} \), and the refractive index \( n = \frac{5}{3} \).

1. Lens Maker’s Formula:
The lens maker’s formula for a lens in air is:

\( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)

2. Sign Convention:
For a double concave lens:
- The first surface is convex from the side of the incident light, so \( R_1 = -20 \, \text{cm} \)
- The second surface is concave from the side of the incident light, so \( R_2 = 40 \, \text{cm} \)
(Note: R₁ is negative and R₂ is positive for a double concave lens)

3. Substituting Values:

\( \frac{1}{f} = \left( \frac{5}{3} - 1 \right) \left( \frac{1}{-20} - \frac{1}{40} \right) \)

\( \frac{1}{f} = \frac{2}{3} \left( \frac{-2 - 1}{40} \right) = \frac{2}{3} \times \left( \frac{-3}{40} \right) = \frac{-6}{120} = \frac{-1}{20} \)

4. Final Calculation:

\( f = -20 \, \text{cm} \)

Final Answer:
The focal length of the lens is \({-20 \, \text{cm}} \).