Power of a Lens Ray Optics - Definition, Formula, FAQs

In this article, we are going to discuss a very important topic of class 10, the power of a lens. A lens is a piece of glass through which light is passed or refracted. Lens are used for fixing vision problems and other instruments that give man a better perspective on day to day things. In the below article, we will discuss the power of the lens for better understanding.

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This Story also Contains

1. What is Lens?
2. Power of a Lens
3. Power of a lens Formula
4. Relation between Refractive Index and Focal Length
5. Solved Examples on Power of Lens

What is Lens?

A lens is a transparent magnifying curved glass through which light rays converge or diverge. Lenses are used in optical instruments such as cameras, microscopes, telescopes, and eyeglasses to manipulate light for various purposes. There are two types of lenses namely Concave lens and Convex lens.

Power of a Lens

We know that a convex lens converges the light rays incident on it but a concave lens diverges the incident light rays. The power of a lens is a measure of its ability to converge or diverge light rays that incident on it .

This ability is based on its focal length. It is found that a convex lens of shorter focal length bends the light rays through larger angles and focuses them closer to the optical center. Similarly, a concave lens of a smaller focal length produces more divergence than a lens of a longer focal length. This shows that the power of the lens is inversely proportional to the focal length of the lens.

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Power of a lens Formula

The power of a lens is mathematically defined as the reciprocal of its focal length.

Si unit of the power of lens is dioptre. It is usually denoted by the letter ‘d’.

Formula:

$$
P=\frac{100}{f}
$$

In SI Units:

$$
P=\frac{1}{f}
$$
Where:
$P=$ Power of the lens (in diopters, $D$ )
$f=$ Focal length of the lens (in centimeters)

Sign Convention for Lens

Convex Lens (Converging): Positive focal length $(f>0)$, hence positive power $(P>0)$.
Concave Lens (Diverging): Negative focal length $(f<0)$, hence negative power $(P<0)$.

Relation between Refractive Index and Focal Length

$$
\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
$$

Where:
$f$ : Focal length of the lens (in meters).
$n$ : Refractive index of the lens material relative to the surrounding medium.
$R_1$ : Radius of curvature of the first lens surface (positive for convex, negative for concave).
$R_2$ : Radius of curvature of the second lens surface (positive for convex, negative for concave).

Optics Formula

Lens Formula

$$
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
$$

Where:
- $f$ : Focal length of the lens.
- $v$ : Image distance.
- $u$ : Object distance.

Magnification for Lenses

$$
M=\frac{h^{\prime}}{h}=\frac{v}{u}
$$

Where:
- $M$ : Magnification in lens
- $h^{\prime}$ : Height of the image.
- $h$ : Height of the object.
- $v$ : Image distance.
- $u$ : Object distance.

Mirror Formula

$$
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
$$

Where:
- $f$ : Focal length of the mirror.
- $v$ : Image distance.
- $u$ : Object distance.

Lens Maker’s Formula

$$
\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
$$

Where:
- $f$ : Focal length.
- $n$ : Refractive index of the lens material.
- $R_1, R_2$ : Radii of curvature of the two lens surfaces.

Solved Examples on Power of Lens

1. A convex lens has a focal length of 0.5 meters. What is its power?
Solution:

$$
P=\frac{1}{f}
$$
Given $f=0.5 \mathrm{~m}$ :

$$
P=\frac{1}{0.5}=+2 \mathrm{D}
$$
The power of the lens is $+\mathbf{2}$ diopters (positive sign indicates it is a convex lens).

2. A lens has a power of -4 D . Find its focal length.
Solution:

$$
P=\frac{1}{f}
$$
Rearranging for $f$ :

$$
f=\frac{1}{P}
$$
Given $P=-4 \mathrm{D}$ :

$$
f=\frac{1}{-4}=-0.25 \mathrm{~m}
$$
The focal length is -0.25 m (negative sign indicates it is a concave lens).

3. A convex lens has a focal length of 50 cm. Find its power.
Solution:

Convert focal length to meters:

$$
\begin{gathered}
f=50 \mathrm{~cm}=0.50 \mathrm{~m} \\
P=\frac{1}{f}=\frac{1}{0.50}=+2 \mathrm{D}
\end{gathered}
$$
The power of the lens is $+\mathbf{2}$ diopters.

3. Two lenses of powers +3 D and -2 D are placed in contact. What is the resultant power?
Solution:

The resultant power is the algebraic sum of the individual powers:

$$
P_{\text {resultant }}=P_{\mathbf{1}}+P_2
$$

Given $P_1=+3 \mathrm{D}$ and $P_2=-2 \mathrm{D}$ :

$$
P_{\text {resultant }}=+3-2=+1 \mathrm{D}
$$

The resultant power is $+\mathbf{1}$ diopter.