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 ?

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$$

Convex lens has a positive focal length ($$f_{\text{convex}} = +f_1$$)

Concave lens has a negative focal length ($$f_{\text{concave}} = -f_2$$)

$$\frac{1}{F} = \frac{1}{|f_{\text{convex}}|} - \frac{1}{|f_{\text{concave}}|}$$

$$\frac{1}{F} = \frac{|f_{\text{concave}}| - |f_{\text{convex}}|}{|f_{\text{convex}}| \cdot |f_{\text{concave}}|}$$

$$F = \frac{|f_{\text{convex}}| \cdot |f_{\text{concave}}|}{|f_{\text{concave}}| - |f_{\text{convex}}|}$$

For the combination to act as a net concave lens, its equivalent focal length must be negative ($$F < 0$$). This occurs when the denominator is less than zero:

$$|f_{\text{concave}}| - |f_{\text{convex}}| < 0 \implies |f_{\text{convex}}| > |f_{\text{concave}}|$$

For the combination to act as a net convex lens, its equivalent focal length must be positive ($$F > 0$$). This occurs when the denominator is greater than zero:

$$|f_{\text{concave}}| - |f_{\text{convex}}| > 0 \implies |f_{\text{convex}}| < |f_{\text{concave}}|$$