A spherical interface lens of radius R separates two media of refractive indices 1 and 1.4 respectively as shown in the figure below

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 ______. [2026]

Smart Achievers · Accepted Answer

(1)

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.4}{v} - \frac{1}{- 4 R} = \frac{1.4 - 1}{R} \Rightarrow v = \frac{28}{3} R$

$m = \frac{μ_{1}}{μ_{2}} \frac{v}{u} = \frac{1}{1.4} \times \frac{28 R}{3 \times 4 R} = \frac{7}{4.2} = 1.66$

Solution

Q.
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 ______. [2026]

1 1.66

2 2.33

3 2.66

4 1.33

Ans.
(1)

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.4}{v} - \frac{1}{- 4 R} = \frac{1.4 - 1}{R} \Rightarrow v = \frac{28}{3} R$

$m = \frac{μ_{1}}{μ_{2}} \frac{v}{u} = \frac{1}{1.4} \times \frac{28 R}{3 \times 4 R} = \frac{7}{4.2} = 1.66$

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