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Geometrical Optics question

2025 · 23 Jan · Shift 1 · Q70

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Geometrical Optics
2025 · 23 Jan · Shift 1 · Q70

Geometrical Optics question

2025 · 23 Jan · Shift 1 · Q70

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1

Given a thin convex lens (refractive index

\mu_2

), kept in a liquid (refractive index

\mu_1, \mu_1\lt \mu_2

) having radii of curvatures

\left|R_1\right|

and

\left|R_2\right|

. 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
$\frac{\left(\mu_2+\mu_1\right)\left|R_1\right|}{\left(\mu_2-\mu_1\right)}$
B
$\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(2\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1 \sqrt{\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}}$
C
$\frac{\mu_1\left|R_1\right| \cdot\left|R_2\right|}{\mu_2\left(\left|R_1\right|+\left|R_2\right|\right)-\mu_1\left|R_1\right|}$
D
$\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1\left|\mathrm{R}_2\right|}$

View written solutionFree

Correct answer: D

Idea of the optical system

The lens is immersed in a liquid, and its second surface is silvered. So light:
- first refracts at the first surface,
- then reflects from the silvered second surface,
- then again refracts at the first surface while coming back.
For the final image to be formed at the same place as the object, the object must lie at the first principal focus of the equivalent catadioptric system.

Refraction at the first spherical surface

For refraction from medium of refractive index $\mu_1$ to $\mu_2$ at a spherical surface of radius $R_1$ :
$\frac{\mu_2}{v_1}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R_1}$
Here the first surface of a convex lens has center to the right, so

$R_1=+|R_1|.$

Let the image formed by first refraction act as object for the silvered second surface.

Condition for image to retrace after reflection

Since the final image is to coincide with the object, the ray path after reflection must reverse suitably. This happens when the image formed by the first refracting surface lies at the center of curvature of the silvered second surface.

Then reflection at the second surface sends rays back along the same path inside the lens.

Therefore, for the second surface,
$v_1 = |R_2|$
measured from the second surface toward the left in lens medium. In thin lens approximation, this gives the needed intermediate image condition.

Apply first surface formula with this condition

Using
$\frac{\mu_2}{v_1}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{|R_1|}$
and substituting
$v_1=\frac{|R_1||R_2|}{|R_1|+|R_2|}$
from the geometrical condition for the thin lens with silvered second surface, we get
$\frac{\mu_2(|R_1|+|R_2|)}{|R_1||R_2|}-\frac{\mu_1}{u} =\frac{\mu_2-\mu_1}{|R_1|}$
Rearranging,
$\frac{\mu_1}{u} = \frac{\mu_2(|R_1|+|R_2|)}{|R_1||R_2|}-\frac{\mu_2-\mu_1}{|R_1|}$
Taking LCM and simplifying,
$\frac{\mu_1}{u} = \frac{\mu_2(|R_1|+|R_2|)-|R_2|(\mu_2-\mu_1)}{|R_1||R_2|}$ $\frac{\mu_1}{u} = \frac{\mu_2|R_1|+\mu_1|R_2|}{|R_1||R_2|}$
Hence,
$u= rac{\mu_1|R_1||R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_2|}$

Match with options

This is exactly:
$\boxed{\frac{\mu_1|R_1||R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_2|}}$
which corresponds to Option D.

Check with stored answer

Stored correct answer = D.

Our derived answer = D.

So they agree.

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