Question

A spherical interface lens of radius $R$ separates two media of refractive indices $1$ and $1.4$ respectively as shown in the figure below. A point source is placed at a distance of $4R$ in front of spherical interface. The magnitude of the magnification of point source image is ————.

• A. 1.66
• B. 2.33
• C. 2.66
• D. 1.33

Hint:

Use the formula for refraction at a spherical surface: (n2/v) - (n1/u) = (n2 - n1)/R and then use the magnification formula m = (n1*v)/(n2*u).

Answer 1

The Correct Option is A

To find the magnification, we first need to determine the image position $v$ using the formula for refraction at a single spherical surface:
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$
Given values from the problem and sign convention:
Refractive index of first medium (air), $n_1 = 1$
Refractive index of second medium, $n_2 = 1.4$
Object distance, $u = -4R$ (distance is measured against incident light)
Radius of curvature, $R = +R$ (the surface is convex towards the rarer medium, so the center of curvature is in the denser medium)

Step 1: Substitute the values into the refraction formula:
$$\frac{1.4}{v} - \frac{1}{-4R} = \frac{1.4 - 1}{R}$$
$$\frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R}$$
Step 2: Solve for $v$:
$$\frac{1.4}{v} = \frac{0.4}{R} - \frac{0.25}{R}$$
$$\frac{1.4}{v} = \frac{0.15}{R}$$
$$v = \frac{1.4}{0.15} R = \frac{140}{15} R = \frac{28}{3} R$$
Step 3: Use the magnification formula for a single spherical surface:
$$m = \frac{n_1 v}{n_2 u}$$
Substituting the values:
$$m = \frac{1 \times (28R/3)}{1.4 \times (-4R)}$$
$$m = \frac{28R/3}{-5.6R} = \frac{28}{3 \times (-5.6)} = \frac{28}{-16.8} = -1.666...$$
The magnitude of magnification is $|m| \approx 1.66$.