Question

A convex lens of focal length $ 10cm $ is placed in contact with a concave lens of focal length $ 20cm $. The focal length of this combination of lenses will be:
(A) $ + 10cm $
(B) $ + 20cm $
(C) $ - 10cm $
(D) $ - 20cm $

Answer 1

Hint: To solve this question, we need to use the formula for the equivalent focal length of the combination of thin lenses which are kept in contact with each other. The given values of the focal length of the lenses are to be substituted in that formula after considering their signs, according to the Cartesian signs convention.

Formula used: The formula which has been used to solve this question is given by
 $\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} + \dfrac{1}{{{f_3}}} +...... + \dfrac{1}{{{f_n}}} $, here $ f $ is the equivalent focal length of the combination of $ n $ lenses having focal lengths $ {f_1},{f_2},{f_3},.........,{f_n} $, which are placed in contact with each other.

Complete step by step answer
Let the focal length of the convex lens be $ {f_1} $, and that of the concave lens be $ {f_2} $.
According to the question, we have the focal length of the convex lens equal to $ 10cm $, and that of the concave lens equal to $ 20cm $.
We know from the Cartesian sign convention that the focal length of a convex lens is positive while that of a concave lens is negative.
So the values of the focal length with proper sign convention are given as
 $\Rightarrow {f_1} = + 10cm $ ………………………….(1)
 $\Rightarrow {f_2} = - 20cm $...........................(2)
Now, we know that the equivalent focal length of the combination of thin lenses in contact is given by the formula
 $\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} + \dfrac{1}{{{f_3}}} +...... + \dfrac{1}{{{f_n}}} $
As there are two lenses of focal lengths $ {f_1} $ and $ {f_2} $, so the equivalent focal length of the given combination is given by
 $\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} $
Substituting (1) and (2) we have
 $\Rightarrow \dfrac{1}{f} = \dfrac{1}{{ + 10}} + \dfrac{1}{{ - 20}} $
 $ \Rightarrow \dfrac{1}{f} = \dfrac{{2 - 1}}{{20}} = \dfrac{1}{{20}} $
Taking the reciprocal we get
 $\Rightarrow f = + 20cm $
Thus, the focal length of the given combination of the lenses is equal to $ + 20cm $.

Hence, the correct answer is option B.

Note
We should not forget to consider the signs of the focal lengths given in the question. Although, only the magnitudes have been given, we have to take the nature of the lenses into consideration to get the correct answer.