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Electromagnetic Waves question

2023 · 30 Jan · Shift 1 · Q67
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  5. /2023 · 30 Jan · Shift 1 · Q67

Electromagnetic Waves question

2023 · 30 Jan · Shift 1 · Q67

JEE MainPhysicsElectromagnetic WavesNumerical+4 / −1
A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 24 W24 \mathrm{~W}24 W. The radius of curvature of hemisphere is 10 cm10 \mathrm{~cm}10 cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ‾\underline{\hspace{2cm}}​× 10−8 N\times~10^{-8} \mathrm{~N}× 10−8 N.
Numerical answer
View written solutionFree

Correct answer: 8

  1. Given data
  • Power emitted by point source: P=24 WP = 24\,\text{W}P=24W
  • Radius of hemispherical reflecting surface: R=10 cm=0.1 mR = 10\,\text{cm} = 0.1\,\text{m}R=10cm=0.1m
  • Inner surface is completely reflecting.
  • Source is placed at the centre of curvature of the hemisphere.

We need the force on the hemisphere due to the light.


  1. Key idea: momentum carried by light

Light carries momentum. If energy EEE is incident, momentum associated is

p=Ecp = \frac{E}{c}p=cE​

For a perfectly reflecting surface, the change in momentum along the normal direction is doubled.

So if power incident is PPP, then radiation force for normal incidence on a perfectly reflecting surface is

F=2PcF = \frac{2P}{c}F=c2P​


  1. Why is incidence normal here?

The source is at the centre of curvature of the hemispherical surface. Hence every ray from the source strikes the hemispherical surface along a radius.

But radius is normal to the spherical surface at that point. Therefore, all rays fall normally on the reflecting inner surface.

So each ray gets reflected back along the same path, and the momentum change is maximum.


  1. Is the entire power incident on the hemisphere?

A point source emits uniformly in all directions. A hemisphere subtends half of the full solid angle around the source.

Hence power falling on the hemispherical surface is half of total power:

Phemisphere=242=12 WP_{\text{hemisphere}} = \frac{24}{2} = 12\,\text{W}Phemisphere​=224​=12W


  1. Force on hemisphere

Since the hemisphere is perfectly reflecting and incidence is normal,

F=2Phemispherec=2×123×108F = \frac{2P_{\text{hemisphere}}}{c} = \frac{2\times 12}{3\times 10^8}F=c2Phemisphere​​=3×1082×12​

F=243×108=8×10−8 NF = \frac{24}{3\times 10^8} = 8\times 10^{-8}\,\text{N}F=3×10824​=8×10−8N

So,

F=8×10−8 NF = 8\times 10^{-8}\,\text{N}F=8×10−8N

Therefore, the required integer is

8\boxed{8}8​


  1. Comparison with stored answer

Stored correct answer is 444.

But the correct force is 8×10−8 N8\times 10^{-8}\,\text{N}8×10−8N, because only half the emitted power falls on the hemisphere, and for perfect reflection at normal incidence the force is 2P/c2P/c2P/c on that incident power.

So the stored answer appears to have missed the factor of 222 due to reflection, or used incorrect incident power logic.

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