Question

Achromatic combination of lenses comprises two lenses of the same material placed $4cm$ apart. If the focal length of one lens is $5cm$, the focal length of the other lens is:
$A)\text{ }2cm$
$B)\text{ 4}cm$
$C)\text{ 6}cm$
$D)\text{ 3}cm$

Answer 1

This problem can be solved by the direct formula for the distance between two lenses in an achromatic combination in terms of the focal lengths of the two lenses. The distance between the two lenses should be equal to half of the sum of the focal lengths of the two lenses.

Formula used: $d=\dfrac{{{f}_{1}}+{{f}_{2}}}{2}$

Complete step by step answer:
We will solve this problem by using the direct formula for the distance between two lenses made of the same material in an achromatic combination. The distance must be equal to half of the sum of the focal lengths of the two lenses. So, let us write this formula.
For an achromatic combination of two lenses of the same material, the distance $d$ between the two lenses is given by
$d=\dfrac{{{f}_{1}}+{{f}_{2}}}{2}$ --(1)
Where ${{f}_{1}},{{f}_{2}}$ are the focal lengths of the two lenses respectively.
Now, let us analyze the question.
The distance between the two lenses in the achromatic combination is given to be $d=4cm$.
The focal length of one of the lenses is ${{f}_{1}}=5cm$.
Let the focal length of the other lens be ${{f}_{2}}$.
Therefore, using (1), we get the focal length of the other lens, that is, ${{f}_{2}}$ as
$4=\dfrac{5+{{f}_{2}}}{2}$
$\therefore 4\times 2=5+{{f}_{2}}$
$\therefore 8=5+{{f}_{2}}$
$\therefore {{f}_{2}}=8-5=3cm$
Hence, we have got the focal length of the second lens as $3cm$.

So, the correct answer is “Option D”.

Note: Students must note that an achromatic combination of two lenses, also called an achromatic doublet, is made up of a convex lens and a concave lens. Hence, the focal length of one of the lenses (convex lens) is positive while the other (concave lens) is negative according to sign convention. However, students must remember that while using formula (1) for calculation, the magnitudes of the focal lengths should be taken without applying sign convention, else they will arrive at a wrong result. It may even happen that the distance between the lenses turns out to be negative which is physically impossible.