Ray Optics

JEE Main level, numerical answer, medium

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Question 131 of 295

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ____\_\_\_\_ cm.

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Answer: 16

The magnification of a thin lens is given by m=ff+um = \frac{f}{f + u}.
Given that the size of the images is equal for two different object positions, the magnitudes of the magnifications must be equal: ∣m1∣=∣m2∣|m_1| = |m_2|.
For a single lens to produce images of the same size at two different object distances, one image must be real and the other must be virtual.
Let u1=−8u_1 = -8 cm and u2=−24u_2 = -24 cm.
Then, m1=−m2m_1 = -m_2 (since one is real and one is virtual).
ff+u1=−ff+u2\frac{f}{f + u_1} = -\frac{f}{f + u_2}
1f−8=−1f−24\frac{1}{f - 8} = -\frac{1}{f - 24}
f−24=−(f−8)f - 24 = -(f - 8)
f−24=−f+8f - 24 = -f + 8
2f=322f = 32
f=16f = 16 cm.
The focal length of the lens is 16 cm.

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