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Electromagnetic Waves question

2021 · 17 Mar · Shift 1 · Q70
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Electromagnetic Waves question

2021 · 17 Mar · Shift 1 · Q70

JEE MainPhysicsElectromagnetic WavesNumerical+4 / −1
If 2.5 ×\times× 10 −-− 6 N average force is exerted by a light wave on a non-reflecting surface of 30 cm2 area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ‾\underline{\hspace{2cm}}​ W/cm2. (Round off to the Nearest Integer) (Assume complete absorption and normal incidence conditions are there)
Numerical answer
View written solutionFree

Correct answer: 25

  1. Use radiation pressure relation for complete absorption

For a normally incident light wave on a non-reflecting (completely absorbing) surface,

Pr=IcP_r = \frac{I}{c}Pr​=cI​

where:

  • PrP_rPr​ = radiation pressure
  • III = intensity (energy flux)
  • c=3×108 m/sc = 3 \times 10^8\ \text{m/s}c=3×108 m/s

Also,

Pr=FAP_r = \frac{F}{A}Pr​=AF​

So,

FA=Ic  ⟹  I=FcA\frac{F}{A} = \frac{I}{c} \implies I = \frac{Fc}{A}AF​=cI​⟹I=AFc​

  1. Substitute the given values

Given:

  • F=2.5×10−6 NF = 2.5 \times 10^{-6}\ \text{N}F=2.5×10−6 N
  • A=30 cm2=30×10−4=3×10−3 m2A = 30\ \text{cm}^2 = 30 \times 10^{-4} = 3 \times 10^{-3}\ \text{m}^2A=30 cm2=30×10−4=3×10−3 m2

Thus,

I=(2.5×10−6)(3×108)3×10−3I = \frac{(2.5 \times 10^{-6})(3 \times 10^8)}{3 \times 10^{-3}}I=3×10−3(2.5×10−6)(3×108)​

  1. Calculate

First, numerator:

(2.5×10−6)(3×108)=7.5×102=750(2.5 \times 10^{-6})(3 \times 10^8) = 7.5 \times 10^2 = 750(2.5×10−6)(3×108)=7.5×102=750

Then,

I=7503×10−3=250000 W/m2I = \frac{750}{3 \times 10^{-3}} = 250000\ \text{W/m}^2I=3×10−3750​=250000 W/m2

  1. Convert to W/cm2^22

Since,

1 m2=104 cm21\ \text{m}^2 = 10^4\ \text{cm}^21 m2=104 cm2

therefore,

I=250000104=25 W/cm2I = \frac{250000}{10^4} = 25\ \text{W/cm}^2I=104250000​=25 W/cm2

  1. About the given time

The time span of 404040 minutes is not needed, because force due to radiation pressure already directly gives intensity.

  1. Final answer

25\boxed{25}25​

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