Physics Question 40 – JEE-MAIN 2026

Substitute the focal lengths into the formula:

Combine the terms to get:

The sign of $F$ determines the nature of the combination. Since $f_{\text{convex}}$ and $|f_{\text{concave}}|$ are positive, the denominator $f_{\text{convex}}|f_{\text{concave}}|$ is always positive. Therefore, the sign of $F$ depends solely on the sign of the numerator $(|f_{\text{concave}}|-f_{\text{convex}})$. If $F>0$, it's a convex lens; if $F<0$, it's a concave lens.

Based on the sign of the numerator:

- If $|f_{\text{concave}}|-f_{\text{convex}}>0⟹|f_{\text{concave}}|>f_{\text{convex}}$, then $\frac{1}{F}>0⟹F>0$. The combination behaves as a convex lens.

- If $|f_{\text{concave}}|-f_{\text{convex}}<0⟹|f_{\text{concave}}|<f_{\text{convex}}$, then $\frac{1}{F}<0⟹F<0$. The combination behaves as a concave lens.

- If $|f_{\text{concave}}|-f_{\text{convex}}=0⟹|f_{\text{concave}}|=f_{\text{convex}}$, then $\frac{1}{F}=0⟹F\to \mathrm{\infty }$. The combination behaves as a plane glass plate.

Comparing these conditions with the given options:

- Option 1: "behaves as concave lens if $|f_{\text{convex}}|>|f_{\text{concave}}|$ ". This matches our second condition ($|f_{\text{concave}}|<f_{\text{convex}}$), where the combination behaves as a concave lens. So, Option 1 is correct.

- Option 4: "Focal length of the lens system will change if the positions of two lenses are interchanged". For lenses in contact, the equivalent focal length formula is commutative, meaning interchanging their positions does not change the equivalent focal length. So, Option 4 is incorrect.