**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:

\[P_{rad}=\dfrac{I}{c^2}\]

\[P_{rad}=1.667\times{10^{-14}}k\cdot\dfrac{g}{m^2}\cdot s\]