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Light with an energy of 25\times 10^{4} Wm^{-2} falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm^{2}, the average force exerted on the surface is:-

  • Option 1)

    1.25\times 10^{-6}N

  • Option 2)

    2.50\times 10^{-6}N

  • Option 3)

    1.20\times 10^{-6}N

  • Option 4)

    3.0\times 10^{-6}N

 

Answers (1)

 

Momentum of EM wave -

p = frac{u}{c}

- wherein

u = Energy of EM wave

c = Speed of light in vacuum

 

 

 For a perfectly reflecting surface 

\Delta p=2p

\Delta p=\frac{2E}{C}\\ Force F=\frac{\Delta p}{\Delta t}=\frac{2\times power}{C}=\frac{2\times 25\times 10^{4}}{3\times 10^{8}}\times 15\times 10^{-4}\\ F=2.5\times 10^{-6}N


Option 1)

1.25\times 10^{-6}N

This solution is incorrect 

Option 2)

2.50\times 10^{-6}N

This solution is correct 

Option 3)

1.20\times 10^{-6}N

This solution is incorrect 

Option 4)

3.0\times 10^{-6}N

This solution is incorrect 

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