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JEE Physics 2023 Question with Solution

A person has been using spectacles of power −1.0 diopter-1.0 \, \text{diopter} for distant vision and a separate reading glass of power 2.0 diopters2.0 \, \text{diopters}. What is the least distance of distinct vision for this person?

  • A

    10 cm10 \, \text{cm}

  • B

    40 cm40 \, \text{cm}

  • C

    30 cm30 \, \text{cm}

  • D

    50 cm50 \, \text{cm}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The person uses spectacles of power −1.0 D-1.0 \, \text{D} for distant vision and a reading glass of power 2.0 D2.0 \, \text{D}.

Find: The least distance of distinct vision.

The lens formula is given by:

1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

From the power of the lens,

P=2 D=2 m−1P = 2 \, \text{D} = 2 \, \text{m}^{-1}

so,

1f=2100 cm−1\frac{1}{f} = \frac{2}{100 \, \text{cm}^{-1}}

Therefore,

f=1002 cm=50 cmf = \frac{100}{2} \, \text{cm} = 50 \, \text{cm}

Now substituting the values in the lens formula:

1v−(−125)=2100\frac{1}{v} - \left(-\frac{1}{25}\right) = \frac{2}{100}

So,

1v+125=150\frac{1}{v} + \frac{1}{25} = \frac{1}{50}

Rewriting,

1v=150−125\frac{1}{v} = \frac{1}{50} - \frac{1}{25} 1v=1−250\frac{1}{v} = \frac{1 - 2}{50} 1v=−150\frac{1}{v} = -\frac{1}{50}

Hence,

v=−50 cmv = -50 \, \text{cm}

Therefore, the least distance of distinct vision is 50 cm50 \, \text{cm}. The correct option is D.

Common mistakes

  • Using the power −1.0 D-1.0 \, \text{D} of the distant-vision spectacles directly to answer the near-point question is incorrect, because the least distance of distinct vision must be inferred from the reading-glass data. Use the reading glass power to determine the person’s near point.

  • Ignoring the sign convention in the lens formula leads to a wrong value of vv. The object for a reading glass is taken at u=−25 cmu = -25 \, \text{cm}, and the final negative sign in v=−50 cmv = -50 \, \text{cm} indicates a virtual image, so the near point magnitude is 50 cm50 \, \text{cm}.

  • Confusing focal length in metres and centimetres gives numerical errors. Since P=2 DP = 2 \, \text{D}, the focal length is 0.5 m0.5 \, \text{m} or 50 cm50 \, \text{cm}. Keep units consistent throughout the substitution.

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