Question

A thin biconvex lens is prepared from the glass (\(\mu = 1.5\)) both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed from the lens at a distance of ________ cm.

• A. 10
• B. 12.5
• C. 13
• D. 13.5

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A silvered lens behaves like a mirror. For the image to coincide with the object, the rays must strike the silvered surface normally (perpendicularly), so they retrace their path. This effectively means the object is placed at the center of curvature of the equivalent mirror system.
Step 2: Key Formula or Approach:
1. Power of equivalent mirror: \(P_{eq} = 2P_L + P_M\).
2. Lens maker's formula: \(P_L = \frac{1}{f_L} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).
3. Focal length of equivalent mirror: \(F_{eq} = -\frac{1}{P_{eq}}\).
Step 3: Detailed Explanation:
For the lens: \(R_1 = 20\) cm, \(R_2 = -20\) cm, \(\mu = 1.5\). \[ P_L = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \times \frac{2}{20} = \frac{1}{20} \text{ cm}^{-1} \] The left surface is silvered, so it acts as a mirror with \(R = 20\) cm. \(P_M = -\frac{1}{f_M} = \frac{2}{R} = \frac{2}{20} = \frac{1}{10} \text{ cm}^{-1}\). Total Power \(P_{eq} = 2(\frac{1}{20}) + \frac{1}{10} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \text{ cm}^{-1}\). Focal length of equivalent mirror \(F_{eq} = -5\) cm. For the image to coincide with the object, the object must be at the center of curvature of this equivalent mirror: \[ u = R_{eq} = 2 \times |F_{eq}| = 2 \times 5 = 10 \text{ cm} \]
Step 4: Final Answer:
The object should be placed at 10 cm.