# Inverse Square Law Formula

• Last Updated : 12 May, 2022

The inverse square law describes the intensity of light in relation to its distance from the source. It states that the intensity of the radiation is inversely proportional to the square of the distance. In other words, the intensity of light to an observer from a source is inversely proportional to the square of the distance between the observer and the source. It is used to calculate the distance or intensity of a particular radiation. As the distance between the source and the observer rises, the intensity of the light from the source decreases.

Formula

The ratio of the intensity of the light source for two different time intervals is equal to the reciprocal of the squares of their respective distances of the object from the source. The intensity and distance are denoted by the symbol I and d respectively. The standard unit of intensity is candelas or lumens while for distance, it is meters. The dimensional formula of intensity and distance is [M1L0T-3] and [M0L1T0].

I ∝ 1/d2

where,

I is the intensity of light,

d is the distance between light source and observer.

Consider light sources of intensity I1 and I2 at distances d1 and d2 respectively. For this scenario, the inverse square formula is given by,

I1/I2 = d22/d12

### Sample Problems

Problem 1. A light source has an intensity of 20 candelas at a distance of 3 m from the object. Calculate its intensity if it is at a distance of 6 m from the object.

Solution:

We have,

I1 = 20

d1 = 3

d2 = 6

Using the formula we get,

I1/I2 = d22/d12

=> 20/I2 = 62/32

=> I2 = 20 (9/36)

=> I2 = 5 candelas

Problem 2. A light source has the intensity of 50 candelas at a distance of 30 m from the object. Calculate its intensity if it is at a distance of 10 m from the object.

Solution:

We have,

I1 = 50

d1 = 30

d2 = 10

Using the formula we get,

=> 50/I2 = 102/302

=> I2 = 50 (900/100)

=> I2 = 50 (9)

=> I2 = 450 candelas

Problem 3. A light source has the intensity of 100 candelas at a distance of 20 m from the object. Calculate its distance if its intensity is 80 candelas.

Solution:

We have,

I1 = 100

I2 = 80

d1 = 20

Using the formula we get,

=> 100/80 = d22/202

=> d22 = (400) (5/4)

=> d22 = 500

=> d2 = 22.4 m

Problem 4. A light source has the intensity of 20 candelas at a distance of 1 m from the object. Calculate its distance if its intensity is 40 candelas.

Solution:

We have,

I1 = 20

I2 = 40

d1 = 1

Using the formula we get,

=> 20/40 = d22/12

=> d22 = 1/2

=> d2 = 0.7071 m

Problem 5. Calculate the value of d1 for d2 = 100 m, I1 = 200 candelas and I2 = 300 candelas.

Solution:

We have,

d2 = 100

I1 = 200

I2 = 300

Using the formula we get,

=> 200/300 = 100/d12

=> d12 = 100 (2/3)

=> d1 = 122.48 m

Problem 6. Calculate the value of I1 for I2 = 150 candelas, d1 = 5 m and d2 = 15 m.

Solution:

We have,

I2 = 150

d1 = 5

d2 = 15

Using the formula we get,

=> I1/100 = 152/52

=> I1 = 32 (100)

=> I1 = 900 candelas

Problem 7. Calculate the value of I2 for I1 = 650 candelas, d1 = 36 m and d2 = 24 m.

Solution:

We have,

I1 = 650

d1 = 36

d2 = 24

Using the formula we get,

=> 650/I2 = 242/362

=> I2 = 650 (1296/576)

=> I2 = 1462.5 candelas

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