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 i s ________ cm.
Physics · Geometrical optics · lens and combination lens,silvered lens
JEE Mains 2026 23 January Evening Shift

Question

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 i s ________ cm.

G1 Physics · Accepted Answer

For a thin lens, magnification can be written as m = f/(f+u), with the Cartesian sign convention. For the two object positions, u₁ = -8 cm, u₂ = -24 cm. The image sizes are equal, so the magnifications have equal magnitudes and opposite signs: m₁ = -m₂. Therefore, f/(f-8) = -f/(f-24). Cancel f: 1/(f-8) = -1/(f-24). Hence, f-24 = -f+8. Therefore, 2f = 32, so f = 16 cm. Hence the answer is 16.