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  • Light of intensity  is incident on a perf

Light of intensity =3 \mathrm{~W} / \mathrm{m}^2 is incident on a perfectly absorbing metal surface of area 1 \mathrm{~m}^2 making an angle of $60^{\circ}$ with the normal. If the force exerted by the photons on the surface is \mathrm{p} \times 10^{-9} (in Newton), then the value of \mathrm{p} is :

Option: 1

5


Option: 2

2


Option: 3

1


Option: 4

3


Answers (1)

best_answer

We will be calculating everything per unit time.
The effective area for the light on the surface is =1 \times \cos 60^{\circ}
Energy falling on the surface = intensity x Effective area =3 \times 1 \times \cos 60^{\circ}
=\frac{3}{2} watt
Momentum carried by the light per sec =3 /(2 \mathrm{c})=5 \times 10^{-9}
So, p=5 \times 10^{-9}

Posted by

Rakesh

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