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(B) The equivalent focal length $F$ of two thin lenses separated by a distance $d$ is given by the formula:$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2

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increases

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(B) The equivalent focal length $F$ of two thin lenses separated by a distance $d$ is given by the formula:$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$Here,$f_1$ is the focal length of the convex lens (positive) and $f_2$ is the focal length of the concave lens (negative).When the lenses are put in contact,the separation distance $d = 0$.Substituting $d = 0$ into the formula,we get:$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$Since $f_1 > 0$ and $f_2  0$).$\frac{1}{F} = \frac{1}{f} - \frac{1}{f'}$As the lenses are brought from a separated state to contact,the term $-\frac{d}{f_1 f_2}$ (which was negative since $f_1 f_2 < 0$) is removed. Effectively,the power of the combination changes such that the focal length increases compared to the separated state.

$A$ convex and a concave lens separated by a distance are then put in contact. How does the focal length of the combination change?

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