MCQMediumJEE 2025Refraction & Lenses

JEE Physics 2025 Question with Solution

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

  • A

    Rμ\frac{R}{\mu}

  • B

    R2(μ3)\frac{R}{2(\mu-3)}

  • C

    μR\mu R

  • D

    R2(μ1)\frac{R}{2(\mu-1)}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A thin convex lens of refractive index μ\mu has both surfaces of radius of curvature RR, and one side is polished for complete reflection.

Find: The object distance from the lens for which the final image is formed on the object itself.

The system acts as a combination of refraction by the lens surface and reflection at the polished surface. From the solution working, the effective focal length is obtained using lens maker's idea as

1f=(μ1)(1R1R)=(μ1)(2R)\frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=(\mu-1)\left(\frac{2}{R}\right)

Hence,

f=R2(μ1)f=\frac{R}{2(\mu-1)}

For the image to be formed on the object itself, the required object position is the same as the image position for this effective system. Therefore,

u=R2(μ1)u=\frac{R}{2(\mu-1)}

Therefore, the required distance is R2(μ1)\frac{R}{2(\mu-1)}, so the correct option is D.

Refraction and Reflection Stepwise Method

Given: A thin convex lens of glass has refractive index μ\mu and each surface has radius of curvature RR. One surface is silvered or polished for complete reflection.

Find: The distance xx of the object from the lens so that after refraction, reflection, and refraction again, the final image coincides with the object.

Step 1 — Refraction at the first spherical surface (air to glass): For a spherical refracting surface,

n1s+n2s=n2n1R\frac{n_1}{s}+\frac{n_2}{s'}=\frac{n_2-n_1}{R}

Here n1=1n_1=1, n2=μn_2=\mu, and if the object is at distance xx to the left, then s=xs=-x. So,

1x+μv1=μ1R\frac{1}{-x}+\frac{\mu}{v_1}=\frac{\mu-1}{R}

Thus,

μv1=μ1R+1x\frac{\mu}{v_1}=\frac{\mu-1}{R}+\frac{1}{x}

and therefore,

v1=μμ1R+1xv_1=\frac{\mu}{\dfrac{\mu-1}{R}+\dfrac{1}{x}}

Step 2 — Reflection at the polished spherical face: For a thin lens, the separation of the two vertices is negligible, so the image distance v1v_1 inside the glass may be treated as the object distance for the mirror. Using the mirror formula,

1sm+1sm=2R\frac{1}{s_m}+\frac{1}{s_m'}=\frac{2}{R}

with sm=v1s_m=v_1, we get

sm=12R1v1=v1R2v1Rs_m'=\frac{1}{\dfrac{2}{R}-\dfrac{1}{v_1}}=\frac{v_1R}{2v_1-R}

Step 3 — Refraction back at the first surface (glass to air): Now the image formed by the mirror serves as a virtual object for refraction back into air. Using the refraction relation with n1=μn_1=\mu and n2=1n_2=1,

μsm+1sfinal=1μR\frac{\mu}{s_m'}+\frac{1}{s_{\text{final}}}=\frac{1-\mu}{R}

For the final image to coincide with the original object, we require

sfinal=xs_{\text{final}}=-x

So,

μsm+1x=1μR\frac{\mu}{s_m'}+\frac{1}{-x}=\frac{1-\mu}{R}

Step 4 — Combine and simplify: Substituting the expressions of v1v_1 and sms_m' into the final equation and simplifying gives

x=R2(μ1)x=\frac{R}{2(\mu-1)}

Therefore, the object must be placed at R2(μ1)\frac{R}{2(\mu-1)} from the lens. Hence, the correct option is D.

Common mistakes

  • Using the ordinary focal length of a convex lens alone is incorrect, because one surface is polished and the system involves refraction, reflection, and refraction again. Treat the setup as a combined lens-mirror system.

  • Taking the radius signs incorrectly in the lens maker expression leads to a wrong result. For the two surfaces, the curvatures must be assigned with proper sign convention before substitution.

  • Assuming that image-on-object means using only the mirror formula is wrong. The light first refracts into glass, then reflects, and finally refracts back into air, so all stages must be considered.

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