Given is a thin convex lens of glass (refractive index \mu ) and each side having radius of curvature R . One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself?

Geometrical Optics PYQ — JEE Main 2025 · 22 Jan · Shift 1 · Q66 with Solution

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Correct answer: B

Interpret the optical system

A thin convex lens has one side silvered/polished for complete reflection. So light from the object:

first passes through the lens,
then reflects from the polished curved surface,
then passes again through the lens.

We need the object position such that the final image is formed back on the object itself.

Focal length of the thin convex lens

For a thin equiconvex lens in air, with radii $R_1=+R$ and $R_2=-R$ ,

\frac{1}{f}= (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) = (\mu-1)\left(\frac{1}{R}-\left(-\frac{1}{R}\right)\right) = \frac{2(\mu-1)}{R}.

Hence,

f = \frac{R}{2(\mu-1)}.

Power of the silvered lens system

A thin lens of power $P=1/f$ with one surface silvered behaves like:

refraction through lens: power $P$ ,
reflection at plane effectively contributes no additional thin-lens separation effect in this approximation,
refraction again through lens: another power $P$ .

Thus total power becomes

P_{\text{eq}} = 2P = \frac{2}{f}.

So equivalent focal length is

F = \frac{1}{P_{\text{eq}}} = \frac{f}{2}.

Substituting $f$ ,

F = \frac{1}{2}\cdot \frac{R}{2(\mu-1)} = \frac{R}{4(\mu-1)}.

But this does not match any option, so we must treat the silvered curved surface properly as a mirror, not just double-pass lens power.

Refraction at first surface + reflection at second surface + refraction again

For a thin lens with second surface silvered, the combination behaves like a concave mirror in the paraxial approximation. The standard equivalent focal length is

\frac{1}{F}=\frac{2}{f}-\frac{1}{f_m},

but here it is easier to use the known result for a silvered thin lens:

P_{\text{eq}} = 2P + P_m,

where $P=1/f$ is lens power and $P_m$ is mirror power of the silvered surface as seen in air.

For the curved reflecting surface of radius $R$ , mirror focal length is

f_m = \frac{R}{2}, \qquad P_m = \frac{2}{R}.

Lens power is

P = \frac{2(\mu-1)}{R}.

Therefore,

P_{\text{eq}} = 2\cdot \frac{2(\mu-1)}{R} + \frac{2}{R} = \frac{4\mu-4+2}{R} = \frac{2(2\mu-1)}{R}.

Hence,

F = \frac{1}{P_{\text{eq}}} = \frac{R}{2(2\mu-1)}.

Condition for image to form on the object itself

For any mirror-like optical system, if the object is placed at the center of curvature, i.e. at distance $2F$ , the image forms back on the object.

Thus required distance is

u = 2F = 2\cdot \frac{R}{2(2\mu-1)} = \frac{R}{2\mu-1}.

Match with options

\boxed{u=\frac{R}{2\mu-1}}

This is Option B.

Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

More from Geometrical Optics

View written solutionFree

Correct answer: B

Interpret the optical system

A thin convex lens has one side silvered/polished for complete reflection. So light from the object:

first passes through the lens,
then reflects from the polished curved surface,
then passes again through the lens.

We need the object position such that the final image is formed back on the object itself.

Focal length of the thin convex lens

For a thin equiconvex lens in air, with radii $R_1=+R$ and $R_2=-R$ ,

\frac{1}{f}= (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) = (\mu-1)\left(\frac{1}{R}-\left(-\frac{1}{R}\right)\right) = \frac{2(\mu-1)}{R}.

Hence,

f = \frac{R}{2(\mu-1)}.

Power of the silvered lens system

A thin lens of power $P=1/f$ with one surface silvered behaves like:

refraction through lens: power $P$ ,
reflection at plane effectively contributes no additional thin-lens separation effect in this approximation,
refraction again through lens: another power $P$ .

Thus total power becomes

P_{\text{eq}} = 2P = \frac{2}{f}.

So equivalent focal length is

F = \frac{1}{P_{\text{eq}}} = \frac{f}{2}.

Substituting $f$ ,

F = \frac{1}{2}\cdot \frac{R}{2(\mu-1)} = \frac{R}{4(\mu-1)}.

But this does not match any option, so we must treat the silvered curved surface properly as a mirror, not just double-pass lens power.

Refraction at first surface + reflection at second surface + refraction again

For a thin lens with second surface silvered, the combination behaves like a concave mirror in the paraxial approximation. The standard equivalent focal length is

\frac{1}{F}=\frac{2}{f}-\frac{1}{f_m},

but here it is easier to use the known result for a silvered thin lens:

P_{\text{eq}} = 2P + P_m,

where $P=1/f$ is lens power and $P_m$ is mirror power of the silvered surface as seen in air.

For the curved reflecting surface of radius $R$ , mirror focal length is

f_m = \frac{R}{2}, \qquad P_m = \frac{2}{R}.

Lens power is

P = \frac{2(\mu-1)}{R}.

Therefore,

P_{\text{eq}} = 2\cdot \frac{2(\mu-1)}{R} + \frac{2}{R} = \frac{4\mu-4+2}{R} = \frac{2(2\mu-1)}{R}.

Hence,

F = \frac{1}{P_{\text{eq}}} = \frac{R}{2(2\mu-1)}.

Condition for image to form on the object itself

For any mirror-like optical system, if the object is placed at the center of curvature, i.e. at distance $2F$ , the image forms back on the object.

Thus required distance is

u = 2F = 2\cdot \frac{R}{2(2\mu-1)} = \frac{R}{2\mu-1}.

Match with options

\boxed{u=\frac{R}{2\mu-1}}

This is Option B.

Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

Geometrical Optics question

Geometrical Optics question

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Geometrical Optics question

Geometrical Optics question

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