Light of intensity -3 W-m2 is incident on a perfectly absorbing metal surface of area 1 m2 making an angle of 60- with the normal- If the force exerted by the photons on the surfae is p- 10-9 in newton find the value of p-

Effective area = Area perpendicular to the light =1×cos 60^∘ Power falling on the surface = intensity × perpendicular area =3× 1×cos 60^∘ =32 W Momentum carr ...

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Standard XII

Physics

Polarisation

Light of inte...

Question

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

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Solution

Effective area = Area perpendicular to the light $= 1 \times cos 60^{\circ}$

Power falling on the surface = intensity $\times$ perpendicular area

$= 3 \times 1 \times cos 60^{\circ} = \frac{3}{2} W$

Momentum carried by the light per second

$= \frac{3 / 2}{velocity of light} = \frac{1.5}{3 \times 10^{8}} = 5 \times 10^{- 9} N$

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Light of intensity =3 W/m2 is incident on a perfectly absorbing metal surface of area 1 m2 making an angle of 60∘ with the normal. If the force exerted by the photons on the surfae is p×10−9 (in newton) find the value of p.

Effective area = Area perpendicular to the light =1×cos60∘ Power falling on the surface = intensity × perpendicular area =3×1×cos60∘=32 W Momentum carried by the light per second =3/2velocity of light=1.53×108=5×10−9 N

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