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JEE Physics 2025 Question with Solution

A convex lens of focal length 30 cm30 \, \text{cm} is placed in contact with a concave lens of focal length 20 cm20 \, \text{cm}. An object is placed at 20 cm20 \, \text{cm} to the left of this lens system. The distance of the image from the lens in cm is _____ .

  • A

    607 cm\frac{60}{7} \, \text{cm}

  • B

    30 cm30 \, \text{cm}

  • C

    15 cm15 \, \text{cm}

  • D

    45 cm45 \, \text{cm}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A convex lens has focal length f1=+30 cmf_1 = +30 \, \text{cm} and a concave lens has focal length f2=−20 cmf_2 = -20 \, \text{cm}. The object is placed at u=−20 cmu = -20 \, \text{cm}.

Find: The distance of the image from the lens system.

For two thin lenses in contact, the equivalent focal length FF is given by

1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}

Substituting the values,

1F=130−120\frac{1}{F} = \frac{1}{30} - \frac{1}{20} 1F=2−360=−160\frac{1}{F} = \frac{2-3}{60} = -\frac{1}{60}

Hence,

F=−60 cmF = -60 \, \text{cm}

Apply Lens Formula

Now apply the lens formula for the equivalent lens system:

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

Here f=−60 cmf = -60 \, \text{cm} and u=−20 cmu = -20 \, \text{cm}. Therefore,

1−60=1v−1−20\frac{1}{-60} = \frac{1}{v} - \frac{1}{-20} 1−60=1v+120\frac{1}{-60} = \frac{1}{v} + \frac{1}{20}

So,

1v=−160−120\frac{1}{v} = -\frac{1}{60} - \frac{1}{20} 1v=−1−360=−460=−115\frac{1}{v} = \frac{-1-3}{60} = -\frac{4}{60} = -\frac{1}{15}

Thus,

v=−15 cmv = -15 \, \text{cm}

The negative sign shows that the image is formed on the same side as the object. Therefore, the distance of the image from the lens is 15 cm15 \, \text{cm}.

The correct option is C.

Common mistakes

  • Using the concave lens focal length as positive is incorrect because, by the Cartesian sign convention, a concave lens has negative focal length. Use f2=−20 cmf_2 = -20 \, \text{cm}, not +20 cm+20 \, \text{cm}.

  • Adding image sign and distance sign carelessly is wrong. The calculation gives v=−15 cmv = -15 \, \text{cm}, but the question asks for the distance from the lens, so the required value is 15 cm15 \, \text{cm}.

  • Applying the lens formula separately to each lens without first finding the equivalent focal length can lead to confusion here. Since the lenses are in contact, first combine them using 1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}.

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