# Power of a Lens

One of the most fascinating ideas of ray optics is the power of a lens. Simply stated, a lens’s capacity to bend light is its strength in Ray Optics. The ability of a lens to refract light that passes through it is proportional to its power. The converging ability of a convex lens is defined by its strength, while the diverging ability of a concave lens is defined by its diverging ability.

The focal length of a lens is inversely proportional to its intensity. High optical strength is aided by a short focal range. Therefore, let’s first discuss some basic concepts related to the power of the lens.

## What is Power of Lens?

The ability of a lens to bend light is really what gives it its power in Ray Optics. The greater a lens’s strength, the greater its ability to refract light passing through it. Power defines the converging ability of a convex lens and the diverging ability of a concave lens. The number of light bends increases as the focal length reduces. As a result, we can assume that the strength of a lens is inversely proportional to its focal length. A short focal length, in essence, leads to high optical power. Thus,

P = 1 / f

Here, P is the power of the lens and f is the focal length of the lens. If the focal length is given in meters (m), the power of the lens is calculated in Diopters (D), as the lens’s unit of power is diopter.

## Power of Lens Formula

Mathematically formula for the power of a lens is defined as:

Power (P) = 1 / Focal Length (f)

P = 1 / f

For instance, if the focal length of a lens is 15 cm, we get 0.15 m when we translate it to meters. Take the reciprocal of 0.15 to get the power of this prism, which is 6.67. As a result, the power of this lens is 6.67 D. This assumes that the power of a lens can be calculated using the radii of curvature of two surfaces and the refractive index of the lens material.

The power of a lens is inversely proportional to its focal length. Therefore, a short focal length lens has more power, whereas a lens of long focal length has less power.

### Important Points on Power of Lens:

- Power of a Convex Lens (Converging Lens) is positive as its focal length is positive.
- Power of a Concave Lens (Diverging Lens) is negative as its focal length is negative.
- Power of a plane glass is 0.

## Power of Lens Formula (Using Refractive Index)

There is a relation between the focal length of the lens and its refractive index, it’s known as** the Lens maker’s formula**. It includes the radius of curvatures of both surfaces. The lens is a part of the hollow sphere of glass, Radius of curvature of the lens is the radius of that sphere. Each lens has two radii of curvature.

Image

According to the Lens maker’s formula:

1/f = (n-1) × (1/R_{1}– 1/R_{2})where,

nis the Refractive index of the materialfis the focal length of the lensRis the radius of curvatures of first surface_{1}Ris the Radius of curvature of second surface_{2}Also the power of a lens is given by,

P = 1 / fNow, from both the above equations:

P = (n-1) × (1/R_{1}– 1/R_{2})This is the required relation between the power and the refractive index of the lens. This formula can be used to find the power of the lens using the refractive index of the material and radius of curvatures of the lens.

## Power of Combination of Lenses

Two or more lenses can be combined in order to increase or decrease the power of lenses. The formula of the combination is simple and is explained below in the article,

Let take two lenses A and B with focal lengths f

_{1}and f_{2}respectively. These two lenses are placed in contact with each other such that their principal axes coincide with each other. An object is placed at O on the principal axis of the combination. Lens A produces an image of the object at E_{1}. This image acts as an object for lens B and the final image is formed at E.PO = u, i.e. object distance for lens A

PE = v, i.e. final image distance

PE

_{1}= V_{1}, i.e. image distance for lens A and object distance for lens B.Using Lens formula on the image formed by lens A:

1/v……(1)_{1}– 1/u = 1/f_{1 }Using Lens formula on the image formed by lens B:

1/v – 1/v……(2)_{1}= 1/f_{2 }Adding equation (1) and (2) as:

1/v – 1/u = 1/f……(3)_{1}+ 1/f_{2 }Replace the combination with a single lens of focal length F, such that the final image is formed at E.

1/v – 1/u = 1/F……(4)Now from equation (1) and (2) as:

1/F = 1/f……(5)_{1}+ 1/f_{2 }where,

Fis the focal length of the combination of lenses A and B.Since,

P = 1 / fTherefore, equation (5) changes to

P = P_{1}+ P_{2}where,

Pis the power of the combination of lenses,Pis the Power of lens A,_{1}Pis the Power of lens B._{2}This is the formula for the combined power of lenses. Proper sign convention needs to be followed when substituting the values of P

_{1}and P_{2}.

## Dioptre Formula

Dioptre Formula calculates the optical power of any lens. Dioptere is the unit for measuring the power of lens. Power of the lens is inversely proportional of the focal length of the lens.

D =1/ fwhere,

Dis the power in dioptresfis the focal length of the lens

## Solved Examples on Power of a Lens

**Example 1: How does the power of a lens change if its focal length is doubled?**

**Solution:**

Power gets halved as Power is inversely proportional to focal length.

**Example 2: What is the power of a convex lens (with sign) of focal length 40cm?**

**Solution:**

Since, Power = 1 / f

Substituting the given values as,

P = 100/40

= 2.5D

Since it’s a convex lens, so power will be positive.

Thus, the power of convex lens is

+2.5D.

**Example 3: Identify the type of lens, and its focal length if its power is 0.2D.**

**Solution:**

Since the focal length, f = 1 / Power (P)

Therefore, substituting the given values in the above expression as:

f = 1 / (0.2D)

= 5 m

Since the power is positive, therefore given lens is a

convex lens.

**Example 4: A convex lens of a focal length of 50 cm is in contact with a concave lens of 20 cm focal length. Find the power of the combination of lenses.**

**Solution:**

For a combination of lenses,

P = P

_{1}+ P_{2}P = 1/f

_{1}+ 1/f_{2}P = 100/50 + 100/(-20) (concave lenses have negative focal length)

=

-3 D

**Example 5: Find the power of a plano-convex lens, when the radius of a concave surface is 10 cm and refractive index is (n) 1.5.**

**Solution:**

Given

R_{1}= ∞, R_{2}= – 10 cm = -0.1 m, n= 1.5

P = (n-1) × (1/R_{1}– 1/R_{2})P = (1.5-1) × (1/∞ – 1/{-0.1})

P = 0.5 × (0 + 10)

P = 5 D

## FAQs on Power of Lens

**Question 1: What is the power of a lens?**

**Answer:**

Power of a lens defines the degree of convergence and divergence of the light that strikes the lens. Thus, variation in power changes the convergence and divergence of the light that passes through the lens.

**Question 2: What is the relation of focal length with the power of the lens?**

**Answer:**

Power of the lens is reciprocal to the focal length of the lens.

**Question 3: What is the SI unit for the power of a lens?**

**Answer:**

SI unit for the power of the lens is Diopter. Its dimension is

m^{-1}.

**Question 4: Which lens has positive power?**

**Answer:**

The power of the Converging lens or Convex lens is always positive.

**Question 5: Sunglasses have curved surfaces but still do not have any power. Why?**

**Answer:**

Sunglasses have two curved surfaces, one is convex and another is concave. Both surfaces are of equal power but of opposite signs, so both the power cancels out each other and resultant power is 0.

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