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Question: A collimated beam of light of flux density \(3kWm^{-2}\) is incident normally on a \(100 mm^{-2}\) c...

A collimated beam of light of flux density 3kWm−23kWm^{-2} is incident normally on a 100mm−2100 mm^{-2} completely absorbing screen. If P is the pressure exerted on the screen and Δp is the momentum transferred to the screen during a 1000s interval, then
A. P=10−3Nm−2P = {10^{ - 3}}N{m^{ - 2}}
B. P=10−4Nm−2P = {10^{ - 4}}N{m^{ - 2}}
C. Δp=10−4Kgms−1\Delta p = {10^{ - 4}}Kgm{s^{ - 1}}
D. Δp=10−5Kgms−1\Delta p = {10^{ - 5}}Kgm{s^{ - 1}}

Explanation

Solution

Momentum is defined as the product of mass and velocity. Force would be exerted on the screen when radiation falls on any surface for an interval of time and there is a change in the momentum of electromagnetic waves.

Complete step by step answer:
Expressing the relation of radiation pressure and intensity of electromagnetic waves i.e. flux density when it is absorbed completely.
P=IcP = \dfrac{I}{c}
Here, PP is the radiation pressure, II is the intensity of electromagnetic waves or and cc is the speed of light.
Substituting 3000 Wm−23000 \, Wm^{-2} flux density and c=3×108ms−1c= 3 \times 10^8 ms^{-1}
⇒P=30003×108=10−4N/m2\Rightarrow P = \dfrac{{3000}}{{3 \times {{10}^8}}} = {10^{ - 4}}N/{m^2}
Here, PP is the radiation pressure which is equal to 10−4N−m−210^{-4} N-m^{-2}
Expressing the relation between force and pressure,
F=P×AF = P \times A
Expressing the relation of momentum, force, and time
⇒Δp=Ft\Rightarrow \Delta p = \dfrac{F}{t}
Here, Δp\Delta p is the change in momentum, FF is the force exerted by the electromagnetic wave on the surface, PP is the pressure and t is the time for which force exerted.
Substituting the value of F in the equation of momentum
Δp=PAt\Delta p = \dfrac{{PA}}{t}
Here, Δp\Delta p is a change in momentum.
Substituting p=10−4Nm−2p= 10^{-4} Nm^{-2} , A=100mm2=10−4m2A= 100 mm^2 = 10^{-4} m^2 and t=1000st=1000 s, we get
⇒Δp=10−4×10−41000=10−5kgm/s\Rightarrow \Delta p = \dfrac{{{{10}^{ - 4}} \times {{10}^{ - 4}}}}{{1000}} = {10^{ - 5}}kgm/s
Here, Δp\Delta p comes out to be 10−510^{-5} kg m/s.

The correct option is B and D.

Note:
Electromagnetic waves carry energy and momentum with them. Different cases of radiation pressure when the beam absorbed or reflected are, If an Electromagnetic wave is absorbed by an object (In case of a black body which absorbs the light falling on it), or it reflects from an object(in case of a polished surface), the wave will transfer momentum to the object. If the wave is incident on the object for a longer duration, more momentum is transferred.