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JEE Physics 2023 Question with Solution

Two convex lenses of focal length 20 cm20 \, \text{cm} each are placed coaxially with a separation of 60 cm60 \, \text{cm} between them. The image of the distant object formed by the combination is at _____ cm from the first lens.

Answer

Correct answer:100

Step-by-step solution

Standard Method

Given: Focal length of each lens, f1=f2=20 cmf_1 = f_2 = 20 \, \text{cm}. Separation between the lenses, d=60 cmd = 60 \, \text{cm}. The object is at infinity.

Find: The distance of the final image from the first lens.

For the first lens, the object is at infinity. Using the lens formula:

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

Since u=∞u = \infty,

120=1v1−1∞\frac{1}{20} = \frac{1}{v_1} - \frac{1}{\infty} 1v1=120\frac{1}{v_1} = \frac{1}{20} v1=20 cmv_1 = 20 \, \text{cm}

So, the first lens forms an image at 20 cm20 \, \text{cm} from itself.

This image acts as the object for the second lens. Hence,

u2=d−v1=60−20=40 cmu_2 = d - v_1 = 60 - 20 = 40 \, \text{cm}

Since this object is on the same side as the incoming light for the second lens, we take

u2=−40 cmu_2 = -40 \, \text{cm}

Now for the second lens,

1f2=1v2−1u2\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}

Substituting the values,

120=1v2−1−40\frac{1}{20} = \frac{1}{v_2} - \frac{1}{-40} 120=1v2+140\frac{1}{20} = \frac{1}{v_2} + \frac{1}{40} 1v2=120−140=140\frac{1}{v_2} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} v2=40 cmv_2 = 40 \, \text{cm}

Therefore, the final image is at a distance

d+v2=60+40=100 cmd + v_2 = 60 + 40 = 100 \, \text{cm}

from the first lens.

Therefore, the image is formed at 100 cm100 \, \text{cm} from the first lens.

Common mistakes

  • Taking the image formed by the first lens as being at the second lens instead of 20 cm20 \, \text{cm} from the first lens is incorrect. First find the image due to the first lens for an object at infinity, then use lens separation to locate the object for the second lens.

  • Using the second lens object distance as +40 cm+40 \, \text{cm} is incorrect under the Cartesian sign convention. The intermediate image lies on the same side as the incoming light for the second lens, so use u2=−40 cmu_2 = -40 \, \text{cm}.

  • Stopping after finding v2=40 cmv_2 = 40 \, \text{cm} is incomplete. That value is measured from the second lens, whereas the question asks for the distance from the first lens. Add the separation 60 cm60 \, \text{cm} to get the final position.

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