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MCQEasyJEE 2025Refraction & Lenses

JEE Physics 2025 Question with Solution

The radii of curvature for a thin convex lens are 10 cm10 \, \text{cm} and 15 cm15 \, \text{cm} respectively. The focal length of the lens is 12 cm12 \, \text{cm}. The refractive index of the lens material is

  • A

    1.21.2

  • B

    1.41.4

  • C

    1.51.5

  • D

    1.81.8

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: For the thin convex lens, R1=+10 cmR_1 = +10 \, \text{cm}, R2=−15 cmR_2 = -15 \, \text{cm}, and f=12 cmf = 12 \, \text{cm}.

Find: The refractive index nn of the lens material.

Use the lens maker's formula:

1f=(n−1)(1R1−1R2)\frac{1}{f} = \left(n - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

Substituting the given values:

112=(n−1)(110−1−15)\frac{1}{12} = \left(n - 1\right)\left(\frac{1}{10} - \frac{1}{-15}\right)

Simplify the bracket:

110+115=330+230=530=16\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}

So,

112=(n−1)16\frac{1}{12} = \left(n - 1\right)\frac{1}{6}

Multiplying both sides by 66:

n−1=612=12n - 1 = \frac{6}{12} = \frac{1}{2}

Therefore,

n=1+12=32=1.5n = 1 + \frac{1}{2} = \frac{3}{2} = 1.5

Therefore, the refractive index of the lens material is 1.51.5. The correct option is C.

Direct Substitution

Given: f=12 cmf = 12 \, \text{cm}, R1=10 cmR_1 = 10 \, \text{cm}, R2=−15 cmR_2 = -15 \, \text{cm}.

Find: μ\mu of the lens material.

Apply directly:

1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) 112=(μ−1)(110−1−15)\frac{1}{12} = (\mu - 1)\left(\frac{1}{10} - \frac{1}{-15}\right) 112=(μ−1)(3+230)=(μ−1)16\frac{1}{12} = (\mu - 1)\left(\frac{3+2}{30}\right) = (\mu - 1)\frac{1}{6}

Hence,

μ−1=12\mu - 1 = \frac{1}{2} μ=32\mu = \frac{3}{2}

Thus, the correct option is C.

Common mistakes

  • Using the wrong sign for R2R_2 is a common mistake. For a convex lens, the second surface has R2<0R_2 < 0 under the standard sign convention. Treating R2R_2 as positive gives an incorrect value of the refractive index.

  • Some students use the lens formula 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u} instead of the lens maker's formula. That formula is for image formation, not for relating focal length to refractive index and radii of curvature.

  • Errors often occur while adding 110\frac{1}{10} and 115\frac{1}{15}. The correct sum is 330+230=16\frac{3}{30} + \frac{2}{30} = \frac{1}{6}, not 125\frac{1}{25} or 215\frac{2}{15}.

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