Question

Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is:

• A. 5 D
• B. 1 D
• C. 20 D
• D. 10 D

Hint:

For combined power of two lenses in contact or separated by a small distance, use the formula: \[ \frac{1}{P} = \frac{1}{P_1} + \frac{1}{P_2} - \frac{d}{P_1 P_2}, \] where \( P_1 \) and \( P_2 \) are the individual powers, and \( d \) is the separation between the lenses.

Answer 1

The Correct Option is D

Step 1: Convert Focal Lengths to Powers
First, we calculate the individual powers of the lenses using the formula: \[ P = \frac{1}{f} \] where \( f \) is in meters.

• For the first lens (\( f_1 = 30 \) cm = 0.3 m): \[ P_1 = \frac{1}{0.3} \approx 3.33 \text{ D} \]
• For the second lens (\( f_2 = 10 \) cm = 0.1 m): \[ P_2 = \frac{1}{0.1} = 10 \text{ D} \]

Step 2: Lens Combination Formula
The equivalent power \( P_{\text{eq}} \) of two lenses separated by distance \( d \) is given by: \[ P_{\text{eq}} = P_1 + P_2 - d \cdot P_1 \cdot P_2 \] Given:

• \( P_1 = 3.33 \) D
• \( P_2 = 10 \) D
• \( d = 10 \) cm = 0.1 m

Step 3: Calculate Equivalent Power
Substitute the values into the formula: \[ P_{\text{eq}} = 3.33 + 10 - (0.1 \times 3.33 \times 10) \] \[ P_{\text{eq}} = 13.33 - 3.33 \] \[ P_{\text{eq}} = 10 \text{ D} \]
Verification
Let's verify using exact fractions:

• \( P_1 = \frac{10}{3} \) D (exact value for 0.3 m)
• \( P_2 = 10 \) D
• Calculation: \[ P_{\text{eq}} = \frac{10}{3} + 10 - \left(0.1 \times \frac{10}{3} \times 10\right) \] \[ P_{\text{eq}} = \frac{10}{3} + \frac{30}{3} - \frac{10}{3} \] \[ P_{\text{eq}} = \frac{30}{3} = 10 \text{ D} \]

Conclusion
The power of the lens combination is 4 (10 D).

Answer 2

Given: \[ f_1 = 30 \, \text{cm}, \quad f_2 = 10 \, \text{cm} \] The formula for the equivalent focal length \(f_{eq}\) when two lenses are separated by a distance \(d\) is given by: \[ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \] Substituting the values: \[ \frac{1}{f_{eq}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{(0.3)(0.1)} \] Simplifying: \[ \frac{1}{f_{eq}} = \frac{1}{0.1} \] Thus, the equivalent focal length: \[ \frac{1}{f_{eq}} = 10 \, \text{D} \] \[ \boxed{\text{Power } = 10 \, \text{D}} \]