Question

A converging lens of focal length 30 cm is placed in contact with another converging lens of unknown focal length, then the possible value for the focal length of the combination is
(A) 15 cm
(B) 60 cm
(C) 36 cm
(D) -12 cm

Answer 1

Hint The power of a lens is the inverse of the focal length of the length. When two lenses are kept in contact, the net power of the combination is the sum of the individual power of the lenses.

Complete step by step answer
We’ve been given that a converging lens is in contact with another converging lens of unknown focal length. Using the relation of power of a lens with its formula, we can determine the focal length of the combination.
Let us assume the focal length of the second unknown lens as${f_2}$. Taking the power of the first lens as ${P_1}$ and the second lens as ${P_2}$, then the power of the lenses can be written as
$\Rightarrow{P_{combo}} = {P_1} + {P_2}$
Since the power of a lens is the inverse of the focal length, we can write
$\Rightarrow\dfrac{1}{{{f_{combo}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
Substituting ${f_1} = 30$, we get
$\Rightarrow\dfrac{1}{{{f_{combo}}}} = \dfrac{1}{{30}} + \dfrac{1}{{{f_2}}}$
$ \Rightarrow {f_{combo}} = \dfrac{1}{{\dfrac{1}{{30}} + \dfrac{1}{{{f_2}}}}}$
In the denominator in the above term, we can say that
$\Rightarrow\dfrac{1}{{30}} + \dfrac{1}{{{f_2}}} > \dfrac{1}{{30}}$ since ${f_2}$ is positive for a converging lens,
So the inverse of the denominator will have a value less than$1/30$, that is ${f_{combo}} < 30\,cm$.
From the options given to us, the only possible choice of the focal length of less than 30 cm is choice (A).

Note
We must be careful in taking into account the sign of the focal length of the lenses. Converging lenses have a positive focal length while the diverging lens has a negative focal length. Since the combination of two converging lenses is also a converging lens, we can rule out option (D) which has a negative focal length.