JEE Main 2026 — Physics Optics

2026mcqeasy

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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Options

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

behaves as concave lens if $|f_{\text{convex}}| < |f_{\text{concave}}|$

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

Focal length of the lens system will change if the positions of two lenses are interchanged

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 $|f_{\text{convex}}| > |f_{\text{concave}}|$

JEE Main 2026 — Physics Optics

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JEE Main 2026 — Physics Optics

Options

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics