Question

Two lenses of power -15D and -5D are in contact with each other. The focal length of the combination:
A. -20cm
B. -10cm
C. +20cm
D. +10cm
E. -5cm

Answer 1

Hint: In this question, the two lenses are in contact with each other. The power for lenses is given, so we first calculate the focal length of the two lenses in cm. After this, we will use the formula for calculating the focal combination of lenses.
Formula used: P = $\dfrac{1}{f}$ and $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$ .

Complete Step-by-Step solution:
In the question, two lenses with negative power are given. It means they are concave lenses.
We know that the power of a lens is given by:
P = $\dfrac{1}{f}$, where ‘f’ is in metre.
Putting the values of power in above formula, the focal length is calculated as follow:
-15 =$\dfrac{1}{{{f_1}}}$
$ \Rightarrow {f_1} = \dfrac{{ - 1}}{{15}} \times 100 = \dfrac{{ - 20}}{3}cm$
Also,
-5 =$\dfrac{1}{{{f_2}}}$
$ \Rightarrow {f_2} = \dfrac{{ - 1}}{5} \times 100 = - 20cm$
Now, we know that the focal length of combination of two lenses with distance between them is zero is given by:
$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
Putting the values of ${f_1}$ and ${f_2}$ in above equation, we get:
$\dfrac{1}{f} = \dfrac{1}{{\dfrac{{ - 20}}{3}}} + \dfrac{1}{{ - 20}} = \dfrac{3}{{ - 20}} + (\dfrac{1}{{ - 20}}) = - \dfrac{3}{{20}} - \dfrac{1}{{20}} = \dfrac{{ - 4}}{{20}}$
f = -5cm.
So, option E is correct.

Note – If the distance between the two lenses is ‘d’ and their focal length be ${f_1}$ and ${f_2}$ then in this case the focal length of combination of lens is given as:
$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$.
Converging (convex) lenses have positive focal lengths, so they also have positive power values. Diverging (concave) lenses have negative focal lengths, so they also have negative power values.