JEE Physics 2026 Question with Solution

Given: The object is placed at two different positions $8 \, \text{cm}$ and $24 \, \text{cm}$ from the lens, and the image sizes are equal.

Find: The focal length $f$ of the thin lens.

The magnification is given by

$$m = \frac{f}{f+u}$$

For the image sizes to be equal at different object positions, one image must be real and the other virtual, so the magnifications are equal in magnitude and opposite in sign.

Let

$$u_1 = -8 \, \text{cm}, \qquad u_2 = -24 \, \text{cm}$$

Then

$$|m_1| = |m_2| \implies \frac{f}{f-8} = -\frac{f}{f-24}$$

So,

$$f - 24 = -(f - 8)$$ $$f - 24 = -f + 8$$ $$2f = 32$$ $$f = 16 \, \text{cm}$$

Therefore, the focal length of the lens is $16 \, \text{cm}$.

Given: The two object distances are $8 \, \text{cm}$ and $24 \, \text{cm}$.

Find: The focal length $f$.

If magnification magnitudes are equal for object distances $d_1$ and $d_2$, then

$$f = \frac{d_1 + d_2}{2}$$

Substituting,

$$f = \frac{8 + 24}{2} = \frac{32}{2} = 16 \, \text{cm}$$

Therefore, the focal length is $16 \, \text{cm}$.