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 ?

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 \(|f_{\text{convex}}| > |f_{\text{concave}}|\)