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
Q. 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 ____ $cm$.
Correct Answer: 16 cm

Explanation

For a thin lens to form images of the same size at two different object positions, one image must be real and the other must be virtual. This situation occurs for a convex lens.

Let the object distance be $u$ and focal length be $f$. Using lens formula:

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

$$ \Rightarrow \frac{1}{v} = \frac{u+f}{uf} $$

Magnification is given by:

$$ m = \frac{v}{u} $$

Substituting $v$:

$$ m = \frac{1}{u}\left(\frac{uf}{u+f}\right) = \frac{f}{u+f} $$

Since the sizes (magnitudes of magnification) of the images are equal but their nature is different:

$$ |m_1| = |m_2| \Rightarrow m_1 = -m_2 $$

Using sign convention:

$$ u_1 = -8\,cm,\quad u_2 = -24\,cm $$

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

Cancelling $f$:

$$ \frac{1}{f-8} = -\frac{1}{f-24} $$

$$ f - 24 = -(f - 8) $$

$$ f - 24 = -f + 8 $$

$$ 2f = 32 $$

$$ f = 16\,cm $$

Final Answer: $$ \boxed{16\text{ cm}} $$

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Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

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