Question

The magnitudes of power of a biconvex lens (refractive index \(1.5\)) and that of a plano-convex lens (refractive index \(1.7\)) are same. If the curvature of the plano-convex lens exactly matches with the curvature of the back surface of the biconvex lens, then the ratio of radii of curvature of the front and back surfaces of the biconvex lens is:

• A. \(5:2\)
• B. \(5:12\)
• C. \(12:5\)
• D. \(2:5\)

Hint:

In lens problems, always apply the lens-maker formula carefully and watch the sign convention for radii.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The power of a lens is given by the Lens Maker's Formula. The problem requires comparing the absolute powers of two different lenses based on their refractive indices and geometric constraints.
Step 2: Key Formula or Approach:
Lens Maker's Formula: \(P = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\).
Step 3: Detailed Explanation:
Let the radii of the biconvex lens be \(R_1\) (front) and \(-R_2\) (back).
Power of the biconvex lens (\(n_1 = 1.5\)):
\[ P_1 = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{-R_2} \right) = 0.5 \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]
For the plano-concave lens (\(n_2 = 1.7\)), one surface is plane (\(R = \infty\)) and the other is concave.
Since its curvature matches the back surface of the biconvex lens, its radius of curvature is \(R_2\).
Power of the plano-concave lens:
\[ P_2 = (1.7 - 1) \left( \frac{1}{-R_2} - \frac{1}{\infty} \right) = 0.7 \left( -\frac{1}{R_2} \right) \]
The magnitudes of the powers are equal: \(|P_1| = |P_2|\)
\[ 0.5 \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = 0.7 \left( \frac{1}{R_2} \right) \]
Divide by 0.1:
\[ 5 \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = \frac{7}{R_2} \]
\[ \frac{5}{R_1} + \frac{5}{R_2} = \frac{7}{R_2} \]
\[ \frac{5}{R_1} = \frac{7}{R_2} - \frac{5}{R_2} = \frac{2}{R_2} \]
\[ \frac{R_1}{R_2} = \frac{5}{2} \]
Step 4: Final Answer:
The ratio of radius of curvature of front and back surface of the biconvex lens is \(5 : 2\).