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Electromagnetic Waves appeared 31 times across 3 years — 3.6% of Physics. This question is from Radiation Pressure.

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The radiation pressure exerted by a 450~W light source on a perfectly reflecting surface placed at 2~m away from it, is :

Solution & Explanation

Related Formula

For a perfectly reflecting surface, the radiation pressure Prad is given by:

Prad = (2I)/(c)

where, I = intensity of the light source, c = speed of light ≈ 3 × 10⁸~m/s.

Core Logic

Let's first calculate the intensity I of the point source at a distance r = 2~m:

I = PowerArea = (P)/(4π r²)

Substitute the given values (P = 450~W and r = 2~m):

I = (450)/(4π × 2²) = (450)/(16π)~W/m²
Step 1: Calculating Radiation Pressure

Now, substitute I into the radiation pressure formula:

Prad = (2 × ((450)/(16π)))/(3 × 10⁸) = (900)/(16π × 3 × 10⁸) Prad = (300)/(16π × 10⁸) = (75)/(4π × 10⁸)~N/m²

Using π ≈ 3.1416:

Prad = (75)/(4 × 3.1416 × 10⁸) = (75)/(12.566) × 10⁻⁸ Prad ≈ 5.968 × 10⁻⁸~Pascals ≈ 6 × 10⁻⁸~Pascals
Pattern Recognition

Remember: Perfectly absorbing surface P = I/c. Perfectly reflecting surface P = 2I/c. Always pay close attention to the surface's properties mentioned in the prompt!

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Reference Study Guides

Electromagnetic Waves

More Electromagnetic Waves Previous-Year Questions — Page 7

Q40 jee_main_2024_31_jan_morning Energy Density Of EM Waves
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ ~Hz and an amplitude of 50 ~Vm⁻¹. The total average energy density of the electromagnetic field of the wave is : [Use ε₀ = 8.85 × 10⁻¹² C² / Nm² ]
  • A. 1.106 × 10⁻⁸ Jm⁻³
  • B. 4.425 × 10⁻⁸ ~Jm⁻³
  • C. 2.212 × 10⁻⁸ ~Jm⁻³
  • D. 2.212 × 10⁻¹⁰ ~Jm⁻³

Solution

Related Formula
Utotal average = (1)/(2)ε₀ E₀²
Core Logic

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so:

Uavg = UE + UB = 2UE = 2 ( (1)/(4)ε₀ E₀² ) = (1)/(2)ε₀ E₀²

Where E₀ is the amplitude of the electric field.

Step 2: Substitution

Given: E₀ = 50 V/m ε₀ = 8.85 × 10⁻¹² C²/(N· m²)

Uavg = (1)/(2) × (8.85 × 10⁻¹²) × (50)² Uavg = (1)/(2) × 8.85 × 10⁻¹² × 2500 Uavg = 1.10625 × 10⁻⁸ J/m³
Chapter Mix

Class 12 Physics: Electromagnetic Waves

More Electromagnetic Waves Questions — jee_main_2025_03_april_morning

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Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)