Question

The focal length of the combination of two convex lenses in contact is $f$ and if they are separated by a distance, then focal length of the combination is ${f_1}$. The correct statement is
(A) $f > {f_1}$
(B) $f = {f_1}$
(C) $f < {f_1}$
(D) $f{f_1} = 1$

Answer 1

To solve this question, we need to use the formula for the combination of lenses for the two cases. From there we will have two equations, which can be compared to get the required result.

Formula Used
1. $\dfrac{1}{{{f_e}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$ where${f_e} =$ focal length of combination of two lenses of focal lengths ${f_1}$ and ${f_2}$ in contact with each other
2. $\dfrac{1}{{{f_e}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$ where ${f_e} =$ focal length of combination of two lenses of focal lengths ${f_1}$ and ${f_2}$ separated by distance $d$

Complete step-by-step solution
As we know, the equivalent focal length of the combination of two lenses in contact is given by the relation
$\dfrac{1}{{{f_e}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$ (1)
Also, we know that the equivalent focal length of the combination of two lenses separated by a distance d is given by the relation
$\dfrac{1}{{{f_e}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$ (2)
Let the focal lengths of the two convex lenses be ${f_a}$ and ${f_b}$.
According to the question, the focal length of the combination of the two convex lenses in contact is $f$
So, from (1) we have
$\dfrac{1}{f} = \dfrac{1}{{{f_a}}} + \dfrac{1}{{{f_b}}}$ (3)
Also, if they are separated by a distance, then the focal length of the combination is${f_1}$. Let $x$ be the distance between the two convex lenses. Then from (2) we have
$\dfrac{1}{{{f_1}}} = \dfrac{1}{{{f_a}}} + \dfrac{1}{{{f_b}}} - \dfrac{d}{{{f_a}{f_b}}}$ (4)
As we know that the focal length of a convex lens is positive .Since both the lenses are convex, so both ${f_a}$ and ${f_b}$ are positive. This means that the product ${f_a}{f_b}$ in the equation (4) is positive.
From (3) and (4), we have
$\dfrac{1}{{{f_1}}} = \dfrac{1}{f} - \dfrac{d}{{{f_a}{f_b}}}$
Since ${f_a}{f_b} > 0$
$\dfrac{1}{{{f_1}}} < \dfrac{1}{f}$
Taking reciprocal, we get
${f_1} > f$
which can also be written as,
$f < {f_1}$

Hence, the correct answer is option C, $f < {f_1}$

Note: Be careful to reverse the sign of the inequality while taking the reciprocal. It is a common mistake of directly taking the reciprocal without reversing the inequality sign. So, the rules of the algebra of inequalities should be kept in mind whenever dealing with inequalities.