A thin convex lens and a thin concave lens… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Optics

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A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses ?

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
behaves as concave lens if $|f_{\text{convex}}| > |f_{\text{concave}}|$
B
behaves as concave lens if $|f_{\text{convex}}| < |f_{\text{concave}}|$
C
behaves as convex lens if $|f_{\text{convex}}| > |f_{\text{concave}}|$
D
Focal length of the lens system will change if the positions of two lenses are interchanged

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Solution

The equivalent focal length $F$ of two thin lenses in contact is given by $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$

For a convex lens, the focal length is positive, so $f_1 = |f_{\text{convex}}|$

For a concave lens, the focal length is negative, so $f_2 = -|f_{\text{concave}}|$

Substituting these values, we get $\dfrac{1}{F} = \dfrac{1}{|f_{\text{convex}}|} - \dfrac{1}{|f_{\text{concave}}|}$

If $|f_{\text{convex}}| > |f_{\text{concave}}|$ , then $\dfrac{1}{|f_{\text{convex}}|} < \dfrac{1}{|f_{\text{concave}}|}$

This implies $\dfrac{1}{F} < 0$ , so $F$ is negative.

A negative equivalent focal length indicates that the combination behaves as a diverging or concave lens.

Interchanging the positions of the lenses does not change the equivalent focal length since $\dfrac{1}{f_1} + \dfrac{1}{f_2} = \dfrac{1}{f_2} + \dfrac{1}{f_1}$ .

Answer: behaves as concave lens if $|f_{\text{convex}}| > |f_{\text{concave}}|$

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