The Intensity Formula expresses the power of a wave transmitted per unit area perpendicular to the direction of propagation, given by \( I = P/A \). This fundamental concept appears in Class 11 Physics (NCERT Chapter 15 — Waves) and Class 12 Physics (Chapter 10 — Wave Optics). It is also a high-weightage topic in JEE Main, JEE Advanced, and NEET. This article covers the intensity formula, its derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Intensity Formulas at a Glance
Quick reference for the most important intensity-related formulas.
- Basic intensity: \( I = \dfrac{P}{A} \)
- Intensity and amplitude: \( I \propto A^2 \)
- Intensity and distance (point source): \( I = \dfrac{P}{4\pi r^2} \)
- Resultant intensity (interference): \( I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta \)
- Maximum intensity: \( I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 \)
- Minimum intensity: \( I_{\min} = (\sqrt{I_1} – \sqrt{I_2})^2 \)
- Intensity and energy density: \( I = u \cdot v \)
What is the Intensity Formula?
The Intensity Formula defines intensity as the radiant power passing through a unit area that is perpendicular to the direction of wave propagation. In simple terms, intensity tells us how much energy a wave delivers to a surface every second. It is a scalar quantity, meaning it has magnitude but no direction.
Intensity is represented by the symbol I. Its SI unit is watts per square metre (W/m²). A stronger wave source radiates more power. Spreading that power over a larger area reduces the intensity at any given point.
In NCERT Class 11 Physics (Chapter 15 — Waves), intensity is introduced in the context of sound waves. In NCERT Class 12 Physics (Chapter 10 — Wave Optics), it reappears in the context of light interference and diffraction. Both contexts rely on the same core Intensity Formula. Understanding this formula is essential for scoring well in CBSE board exams and for tackling wave-based problems in JEE and NEET.
Intensity Formula — Expression and Variables
The fundamental expression for intensity is:
\[ I = \frac{P}{A} \]
For a point source radiating uniformly in all directions, the area is the surface area of a sphere of radius r:
\[ I = \frac{P}{4\pi r^2} \]
Intensity is also related to the amplitude of a wave:
\[ I \propto A^2 \]
And to energy density \( u \) and wave speed \( v \):
\[ I = u \cdot v \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | Intensity of the wave | W/m² (Watts per square metre) |
| P | Power of the source | W (Watts) |
| A | Area through which power is distributed | m² (square metres) |
| r | Distance from point source | m (metres) |
| A (amplitude) | Amplitude of the wave | m (metres) |
| u | Energy density | J/m³ |
| v | Wave speed | m/s |
| δ | Phase difference between two waves | radians (dimensionless) |
Derivation of the Intensity Formula
Consider a point source emitting waves uniformly in all directions with total power P. After time t, the wave energy spreads over the surface of an imaginary sphere of radius r centred at the source.
Step 1: The surface area of a sphere is \( A = 4\pi r^2 \).
Step 2: Intensity is defined as power per unit area. So:
\[ I = \frac{P}{A} = \frac{P}{4\pi r^2} \]
Step 3: This shows that intensity follows an inverse-square law. If the distance doubles, the intensity decreases by a factor of four.
Step 4: Since energy density \( u = \frac{\text{Energy}}{\text{Volume}} \) and energy passes through area A at speed v, the power is \( P = u \cdot v \cdot A \). Dividing by A gives \( I = uv \).
This derivation is directly aligned with NCERT Class 11 Physics, Chapter 15.
Complete Physics Intensity Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Basic Intensity | \( I = P/A \) | P = Power, A = Area | W/m² | Class 11, Ch 15 |
| Point Source Intensity | \( I = P/(4\pi r^2) \) | P = Power, r = distance | W/m² | Class 11, Ch 15 |
| Intensity & Amplitude | \( I \propto A^2 \) | A = amplitude of wave | W/m² | Class 12, Ch 10 |
| Intensity & Energy Density | \( I = u \cdot v \) | u = energy density, v = wave speed | W/m² | Class 11, Ch 15 |
| Resultant Intensity (Interference) | \( I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta \) | I&sub1;, I&sub2; = individual intensities, δ = phase difference | W/m² | Class 12, Ch 10 |
| Maximum Intensity (Constructive) | \( I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 \) | I&sub1;, I&sub2; = individual intensities | W/m² | Class 12, Ch 10 |
| Minimum Intensity (Destructive) | \( I_{\min} = (\sqrt{I_1} – \sqrt{I_2})^2 \) | I&sub1;, I&sub2; = individual intensities | W/m² | Class 12, Ch 10 |
| Inverse Square Law Ratio | \( I_1/I_2 = r_2^2/r_1^2 \) | r&sub1;, r&sub2; = distances from source | Dimensionless ratio | Class 11, Ch 15 |
| Sound Intensity Level | \( L = 10\log_{10}(I/I_0) \) | I&sub0; = 10&sup-12; W/m² (threshold) | dB (decibels) | Class 11, Ch 15 |
| Equal Intensity Sources | \( I_{\max} = 4I_0,\; I_{\min} = 0 \) | I&sub0; = intensity of each source | W/m² | Class 12, Ch 10 |
Intensity Formula — Solved Examples
Example 1 (Class 9–10 Level): Basic Intensity Calculation
Problem: A light bulb radiates 100 W of power uniformly in all directions. Calculate the intensity of light at a distance of 2 m from the bulb.
Given: P = 100 W, r = 2 m
Step 1: Write the intensity formula for a point source:
\[ I = \frac{P}{4\pi r^2} \]
Step 2: Substitute the known values:
\[ I = \frac{100}{4\pi (2)^2} = \frac{100}{4\pi \times 4} = \frac{100}{16\pi} \]
Step 3: Calculate the numerical value:
\[ I = \frac{100}{50.27} \approx 1.99 \text{ W/m}^2 \]
Answer
The intensity of light at 2 m from the bulb is approximately 1.99 W/m².
Example 2 (Class 11–12 Level): Inverse Square Law Application
Problem: The intensity of sound from a loudspeaker is 0.80 W/m² at a distance of 3 m. What will be the intensity at a distance of 9 m from the same source? Assume the source radiates uniformly.
Given: I&sub1; = 0.80 W/m², r&sub1; = 3 m, r&sub2; = 9 m
Step 1: Use the inverse square law ratio:
\[ \frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} \]
Step 2: Rearrange for I&sub2;:
\[ I_2 = I_1 \times \frac{r_1^2}{r_2^2} \]
Step 3: Substitute the values:
\[ I_2 = 0.80 \times \frac{(3)^2}{(9)^2} = 0.80 \times \frac{9}{81} = 0.80 \times \frac{1}{9} \]
Step 4: Compute the result:
\[ I_2 = \frac{0.80}{9} \approx 0.089 \text{ W/m}^2 \]
Answer
The intensity at 9 m is approximately 0.089 W/m². The intensity decreased by a factor of 9, consistent with the inverse square law.
Example 3 (JEE/NEET Level): Interference of Two Coherent Sources
Problem: Two coherent light sources have intensities I&sub1; = 4I and I&sub2; = I. Find the ratio of maximum to minimum intensity in the interference pattern.
Given: I&sub1; = 4I, I&sub2; = I
Step 1: Write the expressions for maximum and minimum intensity:
\[ I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 \]
\[ I_{\min} = \left(\sqrt{I_1} – \sqrt{I_2}\right)^2 \]
Step 2: Calculate \( \sqrt{I_1} \) and \( \sqrt{I_2} \):
\[ \sqrt{I_1} = \sqrt{4I} = 2\sqrt{I}, \quad \sqrt{I_2} = \sqrt{I} \]
Step 3: Substitute into the formulas:
\[ I_{\max} = \left(2\sqrt{I} + \sqrt{I}\right)^2 = \left(3\sqrt{I}\right)^2 = 9I \]
\[ I_{\min} = \left(2\sqrt{I} – \sqrt{I}\right)^2 = \left(\sqrt{I}\right)^2 = I \]
Step 4: Find the ratio:
\[ \frac{I_{\max}}{I_{\min}} = \frac{9I}{I} = 9 \]
Answer
The ratio of maximum to minimum intensity is 9 : 1. This is a classic JEE-pattern problem on wave optics interference.
CBSE Exam Tips 2025-26
- Memorise the inverse square law: Questions in CBSE 2025-26 papers frequently ask students to compare intensities at two different distances. Always write \( I \propto 1/r^2 \) as your first step.
- State the formula clearly: In 3-mark and 5-mark questions, write the formula before substituting. CBSE awards step marks for correct formula writing.
- Do not confuse amplitude and intensity: Intensity is proportional to the square of amplitude (\( I \propto A^2 \)). Doubling the amplitude quadruples the intensity. We recommend practising at least five such ratio problems.
- Learn the interference formulas together: The formulas for \( I_{\max} \) and \( I_{\min} \) always appear as a pair. Memorise both together and practise deriving the ratio \( I_{\max}/I_{\min} \).
- Use consistent SI units: Always convert power to Watts and area to m² before substituting. Unit errors cost marks in CBSE 2025-26 exams.
- Practice decibel problems: The sound intensity level formula \( L = 10\log_{10}(I/I_0) \) appears in Class 11 CBSE exams. Know that the threshold intensity \( I_0 = 10^{-12} \) W/m².
Common Mistakes to Avoid
- Mistake 1 — Using diameter instead of radius: Many students use the diameter of a sphere when applying \( I = P/(4\pi r^2) \). Always use the radius r (half the diameter) in the formula.
- Mistake 2 — Confusing intensity with amplitude: Intensity is proportional to the square of amplitude, not amplitude itself. If amplitude doubles, intensity becomes four times, not two times.
- Mistake 3 — Ignoring the phase difference in interference: Students often forget the \( 2\sqrt{I_1 I_2}\cos\delta \) term in the resultant intensity formula. This term is zero only when \( \delta = 90° \), not in general.
- Mistake 4 — Applying the point-source formula to non-spherical sources: The formula \( I = P/(4\pi r^2) \) applies only to isotropic (uniform all-direction) point sources. For a beam or cylindrical source, the geometry changes.
- Mistake 5 — Mixing up \( I_{\max} \) and \( I_{\min} \) conditions: Constructive interference occurs when \( \delta = 0, 2\pi, 4\pi \ldots \) (even multiples of \( \pi \)), giving \( I_{\max} \). Destructive interference occurs when \( \delta = \pi, 3\pi \ldots \) (odd multiples of \( \pi \)), giving \( I_{\min} \).
JEE/NEET Application of Intensity Formula
In our experience, JEE aspirants encounter the Intensity Formula in at least two to three questions per paper, spanning both JEE Main and JEE Advanced. NEET also tests intensity in the context of sound and light waves. Below are the three most common application patterns.
Pattern 1: Intensity Ratio Using Inverse Square Law
JEE Main frequently asks: “If a source is moved from distance r to 2r, by what factor does intensity change?” The answer uses \( I \propto 1/r^2 \). At double the distance, intensity becomes one-fourth. This pattern tests conceptual understanding of the inverse square law. Our experts suggest practising these as one-line mental calculations.
Pattern 2: Interference Pattern — Ratio of \( I_{\max} \) to \( I_{\min} \)
JEE Advanced and NEET both test the interference formula extensively. A typical question gives the ratio of amplitudes \( a_1 : a_2 \) and asks for \( I_{\max} : I_{\min} \). The key steps are: (1) convert amplitude ratio to intensity ratio using \( I \propto A^2 \), then (2) apply \( I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 \) and \( I_{\min} = (\sqrt{I_1} – \sqrt{I_2})^2 \). In our experience, students who memorise these two formulas together solve such problems in under 90 seconds.
Pattern 3: Sound Intensity Level in Decibels
NEET Biology and Physics sections occasionally include questions on the loudness of sound. The formula \( L = 10\log_{10}(I/I_0) \) converts intensity to decibels. A common question asks for the change in decibel level when intensity doubles or increases tenfold. Knowing that a tenfold increase in intensity raises the level by exactly 10 dB is a key shortcut for NEET 2025.
FAQs on Intensity Formula
For more related Physics formulas, explore our comprehensive guides on the Electric Flux Formula, the Angular Speed Formula, and the Average Acceleration Formula. You can also visit our main Physics Formulas hub for the complete list of NCERT-aligned formula articles. For the official NCERT curriculum reference, visit ncert.nic.in.