Question

Given a thin convex lens (refractive index \( \mu_2 \)), kept in a liquid (refractive index \( \mu_1, \mu_1<\mu_2 \)) having radii of curvature \( |R_1| \) and \( |R_2| \). Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?

• A. \( \frac{\mu_1 |R_1| |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_1|} \)
• B. \( \frac{\mu_1 |R_1| |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_2|} \)
• C. \( \frac{\mu_1 |R_1| |R_2|}{\mu_2 (2|R_1| + |R_2|) - \mu_1 \sqrt{|R_1| |R_2|}} \)
• D. \( \frac{(\mu_2 + \mu_1) |R_1|}{\mu_2 - \mu_1} \)

Hint:

In optics problems involving lenses in different media, remember to account for the refractive indices of the medium and the lens, as well as the curvature of the lens surfaces. This can be crucial for solving for the object distance and focal length.

Answer 1

The Correct Option is B

To solve the problem, we need to determine where an object should be placed on the optic axis such that a real and inverted image is formed at the same location by a thin convex lens with one surface silvered. We will employ the lens-maker's formula and mirror formula in conjunction. The lens functions both as a lens and a mirror, forming a catadioptric system.

1. Begin with the lens-maker's formula for a thin lens: 
   \(\frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)\)
2. Determine the focal length of the lens before silvering, \(f'\): 
   \( \frac{1}{f'} = \left(\frac{\mu_2 - \mu_1}{\mu_1}\right)\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right) \)
3. Once the second surface is silvered, the setup functions as a lens-mirror system. The effective focal length \(F\) of the system is: 
   \(\frac{1}{F} = \frac{2}{f'} + \frac{1}{r}\)
   where \(r = -|R_2|\) due to the mirror's curvature.
4. Substitute the value of \(f'\): 
   \(\frac{1}{F} = \frac{2 \mu_1}{(\mu_2 - \mu_1)} \left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right) - \frac{1}{|R_2|}\)
5. Simplify to find the effective focal length: 
   \(\frac{1}{F} = \frac{\mu_1 \left(2|R_2| - |R_1|\right)}{|R_1| |R_2| (\mu_2 - \mu_1)}\)
6. For the formation of a real and inverted image at the object location (autoconjugate point), the object distance \(u\) should be equal to the effective focal length in magnitude. Therefore, \(u = F\).
7. Simplifying using the effective focal length formula, the object distance is: 
   \(u = \frac{\mu_1 |R_1| |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_2|}\)

This is the correct expression for the object distance for the condition specified. Thus, the object should be placed at this distance for a real and inverted image to form at the same location.