Question

A thin plano-convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is:

• A. 0.15 m
• B. 0.10 m
• C. 0.20 m
• D. 0.25 m

Hint:

In problems involving immersion of lenses in liquids, the effective refractive index is the ratio of the refractive index of the lens material to that of the liquid.

Answer 1

The Correct Option is B

Step 1: Find the effective focal length of the lens. The lens maker's formula is: \[ \frac{1}{f} = \left( \frac{n_{lens}}{n_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Here, $n_{lens} = 1.5$, $n_{medium} = 1.2$, $R_1 = R$ (radius of curvature of the curved surface), and $R_2 = \infty$ (since it is a plano-convex lens). \[ \frac{1}{f} = \left( \frac{1.5}{1.2} - 1 \right) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \left( \frac{1.5 - 1.2}{1.2} \right) \frac{1}{R} = \frac{0.3}{1.2} \frac{1}{R} = \frac{1}{4R} \] Therefore, $f = 4R$. 
Step 2: Account for the silvered surface. Since the plane surface is silvered, it acts as a mirror. The effective focal length $F$ of the lens-mirror system is given by: \[ \frac{1}{F} = \frac{2}{f} + \frac{1}{f_m} \] where $f$ is the focal length of the lens and $f_m$ is the focal length of the mirror. In this case, the plane surface acts as a plane mirror, so $f_m = \infty$, and $\frac{1}{f_m} = 0$. Thus, \[ \frac{1}{F} = \frac{2}{f} \] Step 3: Substitute the known values and solve for R. We are given that the effective focal length $F = 0.2$ m. \[ \frac{1}{0.2} = \frac{2}{4R} = \frac{1}{2R} \] \[ 2R = 0.2 \] \[ R = 0.1 { m} \] Therefore, the radius of curvature of the curved surface of the lens is 0.10 m. The correct answer is (2).