Question

A convex lens of focal length $25m$ is placed coaxially in contact with a concave lens of focal length $20cm$. Determine the power of combination. Will the system be converging or diverging in nature?

Answer 1

Hint: The focal length will be positive in a convex lens but in a concave lens the focal length will be negative. If the focal length is positive then it is converging in nature. If the focal length is negative then it is diverging in nature. In a concave lens the incident ray diverges away from the principal axis but in convex the incident ray converges towards the principal axis. The sign convention is assumed to be positive in the direction of incident light.

Formula Used:
 The focal length equation is
 $\dfrac{1}{f} = \dfrac{1}{o} + \dfrac{1}{i}$
Where, $o$ is the original object
 $i$ is the distance from the image to the lens
 $f$ is focal length
 The power of combination,
 $P = {P_1} + {P_2}$
 ${P_1} = \dfrac{1}{{{f_1}}}$
${P_2} = \dfrac{1}{{{f_2}}}$
Where,
 ${P_1}$ is the power of the lens from focal length ${f_1}$
 ${P_2}$ is the power of the lens from focal length ${f_2}$

Complete step by step answer:
To find the power of the combination:
The combination acts as a convex lens whose focal length is,
$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
${P_1} = \dfrac{1}{{{f_1}}}$
${P_2} = \dfrac{1}{{{f_2}}}$
Calculation of focal length combination,
$ \Rightarrow $ $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
$ \Rightarrow $ $\dfrac{1}{f} = \dfrac{1}{{ + 25}} + \dfrac{1}{{ - 20}}$
$ \Rightarrow $ $\dfrac{1}{f} = - \dfrac{1}{{100}}$
$ \Rightarrow $ $f = - 100cm$
Whenever the power of the lens is joined then it is equal to the algebraic sum of power of its individual lenses; this is known as power of combination.
Calculation of power combination,
$ \Rightarrow $ $P = {P_1} + {P_2}$
$ \Rightarrow $ $P = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
$ \Rightarrow $ $P = \dfrac{1}{{0.25}} + \dfrac{1}{{ - 0.20}}$
$ \Rightarrow $ $P = \dfrac{{100}}{{25}} - \dfrac{{100}}{{20}}$
$ \Rightarrow $ $P = \dfrac{{400 - 500}}{{100}}$
$ \Rightarrow $ $P = \dfrac{{ - 100}}{{100}}$
$ \Rightarrow $ $P = - 1D$
It shows negative.
So the combination of lenses is diverging in nature.

Note: Lenses have short focal lengths but they are more powerful. One dioptre is the power of a lens of focal length one meter. It can be denoted by D. the magnification of image distant objects can be determined by the Focal length. In this problem we should check that all measurements use the same measurement system.