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

We are given a thin convex lens with refractive index \( \mu_2 \), placed in a liquid with refractive index \( \mu_1 \), and the radii of curvature of the lens are \( |R_1| \) and \( |R_2| \). The second surface is silver-polished. We need to find the position of the object on the optic axis that will result in a real and inverted image being formed at the same place. To begin, we use the formula for the focal length of a lens: \[ \frac{1}{f_L} = \left( \frac{\mu_2 - \mu_1}{\mu_1} \right) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \] Now, for this system, the object distance \( u \) is related to the focal length. The lens equation gives us: \[ \frac{1}{f_L} = \frac{1}{u} + \frac{1}{v} \] Where \( u \) is the object distance and \( v \) is the image distance. In this case, the image is formed at the same location as the object, so the object distance is equal to the image distance, \( u = v \). Thus, the object should be placed at the following distance: \[ u = \frac{\mu_1 |R_1| |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_2|} \] This matches option (2).