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Question: Define power of lens. What is its unit? One student uses a lens of focal length \(50cm\) and another...

Define power of lens. What is its unit? One student uses a lens of focal length 50cm50cm and another of −50cm - 50cm. What is the nature of the lens and its power used by each of them?

Explanation

Solution

Power of lens is defined as the inverse of the focal length of the lens where the focal length is measured in meters which is denoted as mm .Focal length is measure of how strongly a lens converges or diverges light and it is half of radius of the lens.
Formula used:
P=1fP = \dfrac{1}{f}
Where,
P= Power
f= Focal length of the lens

Complete step by step answer:
Power of a lens is defined as the ability of the lens to bend the light which is falling on it. As power and focal length are inversely proportional to each other then we can conclude that if the lens has shorter focal length then the lens has the ability to bend the light rays more which means power is more.
We know the SI unit of focal length is meter , as it has an inverse relationship between power the SI unit of power is inverse of meter which is denoted as m−1{m^{ - 1}} .This is also known as diopter and denoted as DD.

As given in the problem,
The focal length of the first lens is 50cm50cm.
As we know the SI unit of focal length is in meters.
Now by converting f1{f_1} into meter we get,
f1=0.5m{f_1} = 0.5m
Where,
1cm=10−2m1cm = {10^{ - 2}}m
As the focal length is positive, the nature of the lens used is a convex lens.
Using the power lens formula we get,
P1=1f1{P_1} = \dfrac{1}{{{f_1}}}
By putting the given value we get,
P1=10.5m{P_1} = \dfrac{1}{{0.5m}}
⇒P1=2m−1\Rightarrow {P_1} = 2{m^{ - 1}}
⇒P1=2D\Rightarrow {P_1} = 2D
Similarly,
The focal length of the second lens is −50cm - 50cm .
As we know the SI unit of focal length is in meters.
Now by converting f2{f_2} into meter we get,
f2=−0.5m{f_2} = - 0.5m
Where,
1cm=10−2m1cm = {10^{ - 2}}m
As the focal length is negative, the nature of the lens used is a concave lens.
P2=1f2{P_2} = \dfrac{1}{{{f_2}}}
By putting the given value we get,
P2=1−0.5m{P_2} = \dfrac{1}{{ - 0.5m}}
⇒P2=−2m−1\Rightarrow {P_2} = - 2{m^{ - 1}}
⇒P2=−2D\Rightarrow {P_2} = - 2D
Hence the solution of this problem
SI unit of power is diopter.
Power of the first lens is 2D2D which is convex in nature and the power of the second lens is −2D - 2D which is concave in nature.

Note:
The focal length of a converging lens is considered to be positive and that of the diverging lens is considered to be negative. Accordingly the power of the lens is calculated, if the focal length is positive then the power of the converging lens is positive and that of the power of the diverging lens is negative.