Question

The focal length of double concave lens having refractive index 1.5 kept in air with two spherical surfaces of radii \( R_1 = 20 \, \text{cm} \) and \( R_2 = 80 \, \text{cm} \) is

• A. 16 cm
• B. 32 cm
• C. 24 cm
• D. 48 cm

Hint:

For a concave lens, use the lensmaker's formula to calculate the focal length based on the radii of curvature and refractive index.

Answer 1

The Correct Option is B

Step 1: Identify the Lens Maker’s Formula
The lens maker’s formula for a lens in air is:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( f \) is the focal length, \( n = 1.5 \), \( R_1 = 20 \, \text{cm} \), and \( R_2 = 80 \, \text{cm} \).

Step 2: Apply the Sign Convention for a Double Concave Lens
For a double concave lens:
- The first surface is concave, so \( R_1 = -20 \, \text{cm} \) (negative as the center of curvature is to the left).
- The second surface is convex to the incident light, so \( R_2 = +80 \, \text{cm} \) (positive as the center of curvature is to the right).

Step 3: Substitute Values into the Formula
- Refractive index term: \( n - 1 = 1.5 - 1 = 0.5 \).
- \( \frac{1}{R_1} = \frac{1}{-20} = -0.05 \, \text{cm}^{-1} \).
- \( \frac{1}{R_2} = \frac{1}{80} = 0.0125 \, \text{cm}^{-1} \).
- \( \frac{1}{R_1} - \frac{1}{R_2} = -0.05 - 0.0125 = -0.0625 \, \text{cm}^{-1} \).
- \( \frac{1}{f} = (0.5) \times (-0.0625) = -0.03125 \, \text{cm}^{-1} \).
- \( f = \frac{1}{-0.03125} = -32 \, \text{cm} \).

Step 4: Analyze the Result
The focal length \( f = -32 \, \text{cm} \) is negative, confirming that the lens is diverging, which aligns with the properties of a double concave lens. The magnitude of the focal length is 32 cm, matching option 3 from the given choices: 16 cm, 24 cm, 32 cm, and 48 cm.

Final Answer: The focal length of the double concave lens is 32 cm (Option 2).