JEE Challenger
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PhyHardAdv 2016 • Paper 1

Image Formation and Reflection in Plano Convex Lens

Consider a plano-convex lens crafted from a material having a refractive index nn. A small object positioned at a distance of 30 cm30\text{ cm} in front of its curved surface yields an image that is twice the size of the object. Moreover, light reflection at the convex surface forms a second faint image located 10 cm10\text{ cm} from the lens. Which among the following options are true?

Options

A

The refractive index of the lens material is 2.52.5

Correct
B

The convex surface has a radius of curvature equal to 45 cm45\text{ cm}

C

The secondary faint image formed is erect and real

D

The focal length associated with the lens is 20 cm20\text{ cm}

Correct

Topics & Concepts

Ray Optics and Optical InstrumentsLenses and MirrorsReflection and Refraction

Step-by-Step Solution

To determine the correct statements, we analyze both the refraction through the plano-convex lens and the reflection at its convex surface.

1. Refraction through the Plano-Convex Lens

The object distance from the curved surface of the lens is u=−30 cmu = -30\text{ cm}.

The magnification produced by a thin lens is given by: m=vum = \frac{v}{u}

We are given that the magnitude of the magnification is 22 (the image size is twice the object size), which gives two possible cases:

  • Case 1: Virtual Image (m=+2m = +2) v=mu=2(−30 cm)=−60 cmv = m u = 2(-30\text{ cm}) = -60\text{ cm} Using the lens formula 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}: 1f=1−60−1−30=160  ⟹  f=+60 cm\frac{1}{f} = \frac{1}{-60} - \frac{1}{-30} = \frac{1}{60} \implies f = +60\text{ cm}

  • Case 2: Real Image (m=−2m = -2) v=mu=−2(−30 cm)=+60 cmv = m u = -2(-30\text{ cm}) = +60\text{ cm} Using the lens formula 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}: 1f=160−1−30=360=120  ⟹  f=+20 cm\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{3}{60} = \frac{1}{20} \implies f = +20\text{ cm}

2. Reflection from the Convex Surface

The secondary faint image is formed by direct reflection of light from the outer convex surface, which acts as a convex mirror.

For a real object placed in front of a convex mirror, the image formed is always virtual and erect. Given that the image distance from the surface is 10 cm10\text{ cm}, we have: u=−30 cm,v=+10 cmu = -30\text{ cm}, \quad v = +10\text{ cm}

Using the mirror formula: 1v+1u=1fm\frac{1}{v} + \frac{1}{u} = \frac{1}{f_m} 110+1−30=1fm  ⟹  1fm=230=115 cm−1\frac{1}{10} + \frac{1}{-30} = \frac{1}{f_m} \implies \frac{1}{f_m} = \frac{2}{30} = \frac{1}{15}\text{ cm}^{-1}

Since fm=R2f_m = \frac{R}{2} for a spherical mirror: 2R=115  ⟹  R=30 cm\frac{2}{R} = \frac{1}{15} \implies R = 30\text{ cm}

Thus, the radius of curvature of the convex surface is R=30 cmR = 30\text{ cm}.

3. Calculating the Refractive Index (nn)

Using the Lens Maker's Formula for the plano-convex lens with R1=+30 cmR_1 = +30\text{ cm} and R2=∞R_2 = \infty: 1f=(n−1)(1R1−1R2)\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) 1f=n−1R\frac{1}{f} = \frac{n - 1}{R}

Substitute R=30 cmR = 30\text{ cm} and f=20 cmf = 20\text{ cm} (from Case 2): 120=n−130\frac{1}{20} = \frac{n - 1}{30} n−1=3020=1.5  ⟹  n=2.5n - 1 = \frac{30}{20} = 1.5 \implies n = 2.5

(Note: Substituting f=60 cmf = 60\text{ cm} would yield n=1.5n = 1.5, but this yields a set of parameters not listed in the options).

Conclusion and Evaluation of Options:

  • Option A: The refractive index of the lens material is n=2.5n = 2.5. (Correct)
  • Option B: The radius of curvature is R=30 cmR = 30\text{ cm}, not 45 cm45\text{ cm}. (Incorrect)
  • Option C: The secondary faint image is formed by a convex mirror, making it erect and virtual (not real). (Incorrect)
  • Option D: The focal length of the lens is f=20 cmf = 20\text{ cm}. (Correct)

Correct Options: A and D

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