MCQEasyJEE 2025Characteristics of EM Waves

JEE Physics 2025 Question with Solution

The radiation pressure exerted by a 450W450 \, \text{W} light source on a perfectly reflecting surface placed at 2m2 \, \text{m} away from it, is :

  • A

    1.5×104Pascals1.5 \times 10^{-4} \, \text{Pascals}

  • B

    00

  • C

    6×105Pascals6 \times 10^{-5} \, \text{Pascals}

  • D

    3×105Pascals3 \times 10^{-5} \, \text{Pascals}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Power of source = 450W450 \, \text{W}, distance = 2m2 \, \text{m}, surface is perfectly reflecting.

Find: Radiation pressure on the surface.

For a perfectly reflecting surface,

Prad=2IcP_{rad} = \frac{2I}{c}

where II is the intensity and c=3×108m/sc = 3 \times 10^8 \, \text{m/s}.

The intensity at distance rr from an isotropic source is

I=Psource4πr2I = \frac{P_{source}}{4\pi r^2}

Substituting the given values,

I=4504π(2)2=45016πI = \frac{450}{4\pi (2)^2} = \frac{450}{16\pi}

So,

I45050.278.95W/m2I \approx \frac{450}{50.27} \approx 8.95 \, \text{W/m}^2

Now the radiation pressure is

Prad=2×8.953×108P_{rad} = \frac{2 \times 8.95}{3 \times 10^8} Prad17.93×1085.97×108N/m2P_{rad} \approx \frac{17.9}{3 \times 10^8} \approx 5.97 \times 10^{-8} \, \text{N/m}^2

Since 1N/m2=1Pa1 \, \text{N/m}^2 = 1 \, \text{Pa},

Prad5.97×108Pa6×108PaP_{rad} \approx 5.97 \times 10^{-8} \, \text{Pa} \approx 6 \times 10^{-8} \, \text{Pa}

the solution works out 6×108Pa6 \times 10^{-8} \, \text{Pa}, but it also states that the correct option is C, whose listed value is 6×105Pascals6 \times 10^{-5} \, \text{Pascals}. This is a discrepancy in the source. Following the solution's declared correct option, the correct option is C.

Detailed Working and Source Discrepancy

Given: A 450W450 \, \text{W} light source and a perfectly reflecting surface at 2m2 \, \text{m}.

Find: The radiation pressure.

  1. Spread the source power over a spherical surface of radius 2m2 \, \text{m}:
A=4πr2=4π(2)2=16πm2A = 4\pi r^2 = 4\pi (2)^2 = 16\pi \, \text{m}^2
  1. Compute the intensity:
I=45016π8.95W/m2I = \frac{450}{16\pi} \approx 8.95 \, \text{W/m}^2
  1. For perfect reflection, pressure is twice the absorbing case:
Prad=2IcP_{rad} = \frac{2I}{c}
  1. Substitute:
Prad=2×8.953×1085.97×108PaP_{rad} = \frac{2 \times 8.95}{3 \times 10^8} \approx 5.97 \times 10^{-8} \, \text{Pa}

Thus the numerical result from the working is 6×108Pa6 \times 10^{-8} \, \text{Pa}.

However, the solution's explicitly says "The Correct Option is C" and option C is 6×105Pascals6 \times 10^{-5} \, \text{Pascals}. Therefore the source contains an internal inconsistency between its calculation and its listed answer. Using the page's declared option label, the answer is recorded as C.

Common mistakes

  • Using Ic\frac{I}{c} instead of 2Ic\frac{2I}{c} for a perfectly reflecting surface. Reflection reverses momentum, so the pressure is doubled compared with complete absorption.

  • Calculating intensity as 450πr2\frac{450}{\pi r^2} or 4502πr2\frac{450}{2\pi r^2}. The source is treated as radiating uniformly in all directions, so the power spreads over 4πr24\pi r^2.

  • Missing the unit conversion that 1N/m2=1Pa1 \, \text{N/m}^2 = 1 \, \text{Pa}. Radiation pressure should be reported in pascals after evaluating the expression.

Practice more Characteristics of EM Waves questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Related questions