Dimensional Analysis of EM Wave Intensity Expression

 

❓ Question

The unit of

$$\frac{\sqrt{2I}}{{\epsilon }_0\mathrm{c}}$$

is to be found, where

• $I$ = intensity of an electromagnetic wave,

• $\varepsilon_0$ = permittivity of free space, and

• $c$ = speed of light.

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Units and dimensions
👉 Electromagnetic waves
👉 Electric field relation.

Main idea:

For electromagnetic waves:

$$\boxed{ I=\frac12\varepsilon_0 c E^2 }$$

Thus:

$$\boxed{ E=\sqrt{\frac{2I}{\varepsilon_0 c}} }$$

So the given expression represents electric field.

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🔷 Step 1 — Use EM Wave Formula 💯

Intensity relation:

$$I=\frac12\varepsilon_0 cE^2$$

Rearranging:

$$E= \sqrt{\frac{2I}{\varepsilon_0 c}}$$

Hence the expression has units of electric field.

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🔷 Step 2 — Unit of Electric Field

Electric field unit:

$$\boxed{ N/C }$$

or equivalently:

$$V/m$$

Among given options:

$$\boxed{ NC^{-1} }$$

is correct.

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🔷 Step 3 — JEE Trap Alert 🚨

❌ $V/m$ ko $Vm$ samajh lena

❌ Square root ke andar dimensions incorrectly simplify karna

❌ Electric field units bhool jaana

Remember:

$$\boxed{ 1\ \text{V/m}=1\ \text{N/C} }$$

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✅ Final Answer

$$\boxed{ NC^{-1} }$$

(Option 3)

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