JEE Main 2026 Jan 23 Shift 2, Physics Q22: Magnification

JEE Main2026Jan 23, Shift 2Physics

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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Solution

Magnification of a thin lens: $m = \frac{f}{f + u}$ , where $u$ is signed object distance (taken negative in Cartesian sign convention, but using magnitudes here for clarity).

Equal image sizes with magnitudes equal mean the magnifications differ in sign:

$\frac{f}{f - 8} = - \frac{f}{f - 24}$

$(f - 24) = - (f - 8) \Rightarrow 2 f = 32 \Rightarrow f = 16$ cm.

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JEE Main 2026 Jan 23 Shift 2, Physics Q22: Magnification

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