Question

What do you mean by power of lens? Write its formula and unit. If $2$ concave lens of power $1D$ and $2D$ are combined with a convex lens of power $1.5D$. What is the resultant focal length of the combination?

Answer 1

Hint:The combination of lens concept is used in many optical instruments wherein the lenses are placed either in contact or with a certain distance between them. The power of the lens combination is found out since the respective power of each lens is given and the total power is found out. The formula relating the power and focal length is applied in order to find the resultant focal length.

Complete step by step answer:
The problem revolves around the concept of the combination of lenses and its corresponding power. In order to find the resultant focal length of the combination of lens used we first need to understand the concept of power in lenses.Power is basically the measure of the degree or the extent to which the lens can converge or diverge the light rays falling on it. This can be measured by taking into consideration the focal length of the lens.

The focal length is the distance a focus from the lens’ optical center. It is the point where all the beams of rays converge or diverge together. The power is measured in terms of the angle formed when the light rays bend after hitting the lens material. The power in lenses is thus defined as the tangent of the angle by which it converges or diverges a beam of light incident on it at a unit distance from its optical center.

To determine the formula for power we need to obtain the relation between power and focal length of the lens as mentioned earlier. When the focal length of the lens is said to be smaller then, its ability to bend light rays is more and hence the angle formed will be more which will in-turn make the power of the lens to be greater. Similarly, when the focal length is said to be larger then, its ability to bend light rays is less and hence the angle formed will be less and the power of the lens is less.

Thus, from this it can be clearly seen that power and focal length is inversely related to each other. Thus the formula for the power of lenses is given by the equation:
$P = \dfrac{1}{f}$
By rearranging the terms we get:
$f = \dfrac{1}{P}$ -----------($1$)
The unit of power is measured in Diopters. This unit Dioptre is defined as the unit of power of the lens when the focal length is said to be in one meter. It is denoted by ’$D$’.Now, we come to the actual problem based on the combination of lenses. The combination of lenses is done in order to magnify an image making it clearer to the eye and also done to increase the sharpness of the image by correspondingly reducing the defects and the aberrations due to lenses.

The combination of lenses occurs when two or more thin lenses are placed together making contact with each other or placed at a distance between each other. We consider that the combination of lenses taken over here are placed together without any distance between them. The combination of lenses produces a final image after light is refracted through all the lenses that are combined and this is the concept of equivalence lenses or combined lenses.

We are asked to find out the resultant focal length of the combination of lenses but for that, we first need to find the equivalence power of the combined lenses. The power for the combination of two or more lenses is given as the sum of the individual powers of each of the lenses that are combined which is known as the equivalence power which will be obtained. Hence as per this concept the formula for equivalence power is given as:
$P = {P_1} + {P_2} + .... + {P_n}$ --------($2$)

The question mentions that two concave lenses of power $1D$ and $2D$respectively are combined with a convex lens of power $1.5D$. Here, the combination of lenses and its corresponding results \[1\]nt power is all thin lenses. Since, there are three thins lens that are going to be combined with each other equation ($2$) becomes:
Since, $n = 3$ we get:
$P = {P_1} + {P_2} + {P_3}$ ----------($3$)

However, since the focal length of a concave lens is said to be negative the power of the concave lens is also said to be negative as seen from the relation given in equation ($1$). Thus, the powers of the two concave lenses after applying the necessary sign conventions are $ - 1D$ and $ - 2D$ respectively. The power of a convex is however always positive. Hence the data that we are given is:

Given, ${P_1} = - 1D$, ${P_2} = - 2D$ and ${P_3} = + 1.5D$. This is after applying sign conventions. Hence we substitute these values in equation ($3$) to get:
$P = - 1 + \left( { - 2} \right) + 1.5$
$ \Rightarrow P = - 1 - 2 + 1.5$
On further simplification we get:
$P = - 3 + 1.5$
$ \Rightarrow P = - 1.5D$
Here, $P$ denotes the equivalence power or the total power of the combination of the lenses. In order to find the resultant focal length after the combination we need to substitute this equivalent power in the equation (\[1\]).

Hence we get:
${f_{res}} = \dfrac{1}{{{P_{eq}}}}$
By substituting the equivalent power we get:
${f_{res}} = \dfrac{1}{{ - 1.5}}$
$ \Rightarrow {f_{res}} = - \dfrac{1}{{1.5}}$
${f_{res}} = - 0.666D$
When we approximate it to two significant figures we get:
$\therefore {f_{res}} \approx - 0.67D$
Hence this is the resultant of the focal length and the overall combination of the lens wherein the combination acts concave in nature corresponding to the negative sign of the focal length.

Note:There is no separate formula for calculating the equivalence power in case of convex lenses and concave lenses separately. The equivalence power formula applies for all lenses coming under the thin lens category, that is, the combination can also include concave and convex lenses together which can be seen in the above solution to the problem. This is a common misconception with respect to problems relating the power of combination of lenses.