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JEE Physics 2023 Question with Solution

A point object, 'O' is placed in front of two thin symmetrical coaxial convex lenses L1L_1 and L2L_2 with focal lengths of 24 cm24 \, \text{cm} and 9 cm9 \, \text{cm} respectively. The distance between the two lenses is 10 cm10 \, \text{cm}, and the object is placed 6 cm6 \, \text{cm} away from lens L1L_1 as shown in the figure. The distance between the object and the image formed by the system of two lenses is _____ cm\text{cm}.

A point object O is placed to the left of convex lens L1, with convex lens L2 to its right. The object is 6 cm from L1 and the two lenses are separated by 10 cm along a common principal axis.
  • A

    8 cm8 \, \text{cm}

  • B

    18 cm18 \, \text{cm}

  • C

    12 cm12 \, \text{cm}

  • D

    34 cm34 \, \text{cm}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Focal lengths are f1=24 cmf_1 = 24 \, \text{cm} and f2=9 cmf_2 = 9 \, \text{cm}. The object is 6 cm6 \, \text{cm} to the left of L1L_1, and the distance between the lenses is 10 cm10 \, \text{cm}.

Find: The distance between the object and the final image formed by the two-lens system.

For the first lens, using the lens formula shown in the solution:

1v+16=124\frac{1}{v} + \frac{1}{6} = \frac{1}{24} 1v=124−16=−18\frac{1}{v} = \frac{1}{24} - \frac{1}{6} = -\frac{1}{8} v=−8 cmv = -8 \, \text{cm}

So the image formed by L1L_1 is 8 cm8 \, \text{cm} to the left of the first lens.

This image acts as the object for the second lens. Hence its distance from L2L_2 is

8+10=18 cm8 + 10 = 18 \, \text{cm}

For the second lens:

1v+118=19\frac{1}{v} + \frac{1}{18} = \frac{1}{9} 1v=19−118=118\frac{1}{v} = \frac{1}{9} - \frac{1}{18} = \frac{1}{18} v=18 cmv = 18 \, \text{cm}

Therefore, the final image is 18 cm18 \, \text{cm} to the right of L2L_2.

Hence the distance between the original object and the final image is

6+10+18=34 cm6 + 10 + 18 = 34 \, \text{cm}

Therefore, the distance between the object and the image is 34 cm34 \, \text{cm}. The correct option is D.

Common mistakes

  • A common mistake is taking the object distance for the second lens as 10−8=2 cm10 - 8 = 2 \, \text{cm}. This is wrong because the first image lies 8 cm8 \, \text{cm} to the left of L1L_1, so its distance from L2L_2 is 8+10=18 cm8 + 10 = 18 \, \text{cm}. Always measure the new object distance from the second lens carefully.

  • Another mistake is ignoring the negative sign in v=−8 cmv = -8 \, \text{cm} for the first lens. This sign shows that the first image is on the same side as the object. If this is missed, the position of the intermediate image and the final answer both become incorrect.

  • Students may also add only lens separation and final image distance, forgetting the initial 6 cm6 \, \text{cm} object distance from L1L_1. The question asks for the distance between the original object and the final image, so the full distance is 6+10+186 + 10 + 18.

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