Find the average radiation pressure (Pascal and atmospheric pressure) of:
- the part that completely absorbs the ground.
- the part that completely reflects the ground.
This question aims to find the average radiation pressure. Radiation pressure is actually mechanical pressure that is exerted on any surface caused by the exchange of momentum between an object and an electromagnetic field.
Expert Answer
(a) The average momentum density is calculated by dividing intensity to the square of the speed of light
\[P_{avg}=\dfrac{Light\: of\: intensity (I)}{Speed\: of \: light (c)^2}=\dfrac{I}{c^2}\]
Plug the values in the equation above:
\[P_{avg}=\dfrac{(2500\dfrac{W}{m^2})}{(3\times{10^{8}}\dfrac{m}{s})^2}\]
\[P_{avg}=2.78\times{10^{-14}}k\cdot\dfrac{g}{m^2}\cdot s\]
(b) $F$ is the unit area force that a wave exerts and radiation pressure is represented by $P_{rad}$ and it is the average value of $\dfrac{dP}{dt}$ divided by the area.
\[Light\: of\: intensity (I)=2500\dfrac{W}{m^2}\]
\[Speed\: of \: light (c)= 3\times10^8 \dfrac{m}{s}\]
Radiation pressure is given by equation:
\[P_{rad}=\dfrac{Light\: of\: intensity}{Speed\: of \: light}=\dfrac{I}{c}\]
Substitute values in the above equation:
\[P_{rad}=\dfrac{I}{c}=\dfrac{2500\dfrac{W}{m^2}}{3\times10^8 \dfrac{m}{s}}\]
\[P_{rad}=8.33\times{10^{-6}}\: Pa\]
The radiation pressure in atmosphere is given as:
\[P_{rad}=(8.33\times{10^{-6}}\:Pa)\times(\dfrac{1 atm}{1.103\times{10^{5}}\:Pa})\]
\[P_{rad}=8.23\times{10^{-11}}\:atm\]
(c) The radiation pressure for the totally reflected light is calculated as:
\[P_{rad}=\dfrac{2\times Light\: of\: intensity (I)}{Speed\: of \: light (c)}=\dfrac{2I}{c}\]
Substitute values in the above equation to find radiation pressure for the totally reflected light:
\[P_{rad}=\dfrac{2I}{c}=\dfrac{2(2500\dfrac{W}{m^2})}{3\times{10^{8}}\dfrac{m}{s}}\]
\[P_{rad}=16.66\times{10{-6}}\:Pa\]
Atmospheric radiation pressure is calculated by:
\[P_{rad}=(16.66\times{10{-6}}\:Pa)\times(\dfrac{1\:atm}{1.1013\times{10^{5}}\:Pa})\]
\[P_{rad}=1.65\times{10^{-10}}\:atm\]
Numerical Results
(a) The average momentum density in the light at the floor is:
\[P_{avg}=2.78\times{10^{-14}}k\cdot\dfrac{g}{m^2}\cdot s\]
(b) The radiation pressure in atmosphere for a totally absorbing section of the floor is:
\[P_{rad}=8.23\times{10^{-11}}\:atm\]
(c) The radiation pressure in the atmosphere for a totally reflecting section of the floor is:
\[P_{rad}=1.65\times{10^{-10}}\:atm\]
Example
At NASA’s Jet Propulsion Laboratory’s $25$-foot space simulator facility, a series of overhead arc lamps can generate a light intensity of $1500 \dfrac {W} {m ^ 2} $ on the facility floor. (This simulates the intensity of sunlight near the planet Venus.)
Find the average radiation pressure (Pascal and atmospheric pressure) of:
– the part that completely absorbs the ground.
– the part that completely reflects the ground.
– Calculate the average momentum density (momentum per unit volume) of light on the ground.
This example aims to find the average radiation pressure and average momentum density in the light on the floor.
(a) “F” is an average force per unit area that a wave exerts and the radiation pressure is represented as $P_{rad}$ and it is the average value of $\dfrac{dP}{dt}$ divided by the area.
\[Light\: of\: intensity (I)=1500\dfrac{W}{m^2}\]
\[Speed\: of \: light (c)= 3\times10^8 \dfrac{m}{s}\]
Radiation pressure is given by equation:
\[P_{rad}=\dfrac{I}{c}\]
\[P_{rad}=5\times{10^{-6}}\: Pa\]
Atmospheric radiation pressure is given as:
\[P_{rad}=4.93\times{10^{-11}}\:atm\]
(b) The radiation pressure for the totally reflected light is calculated as:
\[P_{rad}=\dfrac{2I}{c}\]
Substitute values in the above equation to find radiation pressure for the totally reflected light:
\[P_{rad}=1\times{10{-5}}\:Pa\]
\[P_{rad}=9.87\times{10^{-11}}\:atm\]
(c) The average momentum density represents the intensity divided by the square of the speed of light:
\[P_{rad}=\dfrac{I}{c^2}\]
\[P_{rad}=1.667\times{10^{-14}}k\cdot\dfrac{g}{m^2}\cdot s\]