Intensity Formula

Occasionally people can hear a single leaf fall to the ground in a peaceful forest. When people are in bed at night, they may hear our blood rushing through our ears. When a passing motorist’s radio is turned up, though, a person can’t hear what the person in the car next to them is saying. These examples demonstrate the loudness of sounds and, as a result, the intensity of sound. These are also linked to the source’s energy vibrations. Sound intensity is the relevant physical quantity, and this concept holds true for all noises, whether heard or not. With examples, we’ll go over sound intensity and the intensity formula in this post. Let’s take a look at the concept,

Intensity Formula

The quantity of energy carried by a wave per unit time across a unit area surface is known as intensity. It is equal to the energy density multiplied by the wave speed. Watts per square meter is the most used measurement unit. The power and amplitude of a wave will determine its intensity. I is the symbol for intensity. The formula for calculating intensity is as follows:

I = P/A

Where,

I = Intensity,

P = Power

A = Area of cross Section

Dimension of Intensity

The Intensity dimensional formula is as follows: [M¹L⁰T^-3]

Derivation of Intensity dimensional formula

Intensity = Power / Area

Here, Dimensional formula of area is [M⁰L²T⁰]

Power = Work / Time

And, Work = Force × displacement = M × a × displacement

So, Dimensional formula of Work is [M] × [M⁰L¹T^-2] × [L] = [M¹L²T^-2]

Therefore, Power = [M¹L²T^-2] × [M⁰L⁰T¹]^-1 = [M¹L²T^-3]

Intensity = [M¹L²T^-3] × [M⁰L²T⁰]^-1

∴ Dimensional formula of Intensity is [M¹L⁰T^-3]

Sample Problems

Question 1: Calculate the intensity of power is 30 KW wave. The surface’s cross-section is 22 × 10⁶ square meters in size.

Solution:

Given: P = 30 KW, A = 22 × 10⁶ m²

Find: I

I = P/A

∴ I = (30 × 10³)/22 × 10⁶)  

∴ I = 1.36 × 10^-3 W/m²

Question 2: Determine the power of a wave with an intensity of 63 × 10^-5 Watt per square meter. The cross-sectional area is 100 square meters.

Solution:

Given: I = 63 × 10^-5 W/m², A = 100 square meter

Find: P

I = P/A

∴ P = I × A

∴ P = 63 × 10^-5 × 100

∴ P = 63 × 10^-3 W

Question 3: What is the intensity of light incident normal to a 21 cm radius circular surface from a 67 W light source?

Solution:

Given: r = 21 cm = 21 × 10^-2 cm, P = 67 W

Find: I

I = P/A

∴ I = P/πr²

∴ I = 67/3.14 × (21 × 10^-2)²

∴ I = 67 / 3.14 × 441 × 10^-4

∴ I = 67 / 1384.74 × 10^-4

∴ I = 0.04838 × 10⁴ W/m²

Question 4: If the intensity is 45 × 10^-4 Watt per square meter and the power is 33 KW, calculate the area of the cross-section.

Solution:

Given: I = 45 × 10^-4 W/m², P = 33 KW

Find: A

I = P/A

∴ A = P/I

∴ A = 33 × 10³ / 45 × 10^-4

∴ A = 0.733 × 10⁷ m²

Conceptual Questions

Question 1: Define Intensity.

Answer:

The quantity of energy carried by a wave per unit time across a unit area surface is known as intensity.

Question 2: What is the Dimension of Intensity?

Answer:

Dimensional formula of Intensity is [M¹L⁰T^-3].