Radiation Pressure MCQ - Practice Questions with Answers

Consider a point source of light emitting photons. And we want to find the number of Photons (n) emitted by this point source per second.

let the wavelength of light emitted by this $=\lambda$ and
the power of the soulce as P (in Watt or $\mathrm{J} / \mathrm{s}$ )
As we know the energy of each photon is given by

$$
E=h \nu=\frac{h c}{\lambda}(\text { in Joule })
$$

where
where $\mathrm{c}=$ Speed of light, $\mathrm{h}=$ Plank's constant $=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{sec}$
$\nu=$ Frequency in $\mathrm{Hz}, \lambda=$ Wavelength of light.
or we can write the energy of each photon as

$$
E=\frac{12400(\mathrm{eV})}{\lambda\left(A^0\right)}
$$

Then ( $n=t$ the number of photons emitted per second) is given as

$$
n=\frac{\text { Power of source }\left(W \text { or } \frac{J}{\sec }\right)}{\text { Energy of each photon }(J)}=\frac{P}{E}=\frac{P}{\frac{h c}{\lambda}}=\frac{P \lambda}{h c}\left(\sec ^{-1}\right)
$$
 

The intensity of light (I) : The intensity of any quantity is defined as that quantity per unit area.

So here, light energy (or radiation ) crossing per unit area normally per second is called intensity of light energy (or radiation ).

And the intensity I is given as

$$
I=\frac{E}{A t}=\frac{P}{A} \quad\left(\text { where } \frac{E}{t}=P=\text { radiation power }\right)
$$

Its unit is $W / m^2$ or $\frac{J}{m^2 * \sec }$
The intensity of light due to a point isotropic source:
Isotropic source means it emits radiation uniformly in all directions.
So The intensity I due to a point isotropic source at a distance $r$ from it is given as

$$
I=\frac{P}{4 \pi r^2} \Rightarrow \text { i.e } I \propto \frac{1}{r^2}
$$
 

The photon flux $(\phi)$ is defined as the number of photons incident on a normal surface per second per unit area.

As we know $n$ ( the number of photons emitted per second) is given as

$$
n=\frac{\text { Power of source }\left(W \text { or } \frac{J}{\sec }\right)}{\text { Energy of each photon }(J)}=\frac{P}{E}\left(\sec ^{-1}\right)
$$

Similarly intensity I is given as

$$
I=\frac{P}{A}
$$

So The photon flux $\left({ }^\phi\right)$ is given as ratio of Intensity (I) to Energy of each photon

$$
\begin{aligned}
& \phi=\frac{\text { Intensity }}{\text { Energy of each photon }}=\frac{I}{E}=\frac{n}{A} \\
& \text { or } \phi=\frac{I}{E}=\frac{I \lambda}{h c}
\end{aligned}
$$

- The photon flux ${ }^\phi{ }^\phi$ ) due to a point isotropic source:

The photon flux $\left({ }^\phi\right)$ due to a point isotropic source at a distance $r$ from it is given as

$$
\phi=\frac{\text { number of photon per sec }}{\text { sur face area of sphere of radius } r}=\frac{n}{4 \pi r^2}
$$
 

Radiation pressure/force- When photons fall on a surface they exert a pressure/force on the surface. The pressure/force experienced by the surface exposed to the radiation is known as Radiation pressure/force.

As we know

$\mathrm{n}=$ Number of emitted photons per sec is given as

$$
n=\frac{P}{E}=\frac{P}{h \nu}=\frac{P \lambda}{h c}
$$
 where $E=$ The energy of each photon and Momentum of each photon is given as $p=\frac{E}{c}=\frac{h}{\lambda}$

And we know the force is given as rate of change of momentum.
I.e For each photon $F=\frac{d p}{d t}$ and for n photons per sec $F=n(\Delta p)$

For a black body, we get $100 \%$ absorption or $a=1$
i.e for this surface $100 \%$ of the photon will be absorbed

$$
|\Delta p|=\left|0-p_i\right|=\frac{h}{\lambda}
$$

So Force is given as

$$
F=n(\Delta p)=\frac{P \lambda}{h c} * \frac{h}{\lambda}=\frac{P}{c}
$$
 

where P=Power

$$
I=\frac{P}{A} \Rightarrow P=I A
$$

So Force is given as $F=\frac{P}{c}=\frac{I A}{c}$
and radiation pressure is given as

$$
\text { Pressure }=\frac{F}{A}=\frac{I}{c}
$$

i.e For black body,

$$
\begin{aligned}
& F=\frac{P}{C} \\
& \text { Pressure }=\frac{I}{c}
\end{aligned}
$$

- For perfectly reflecting surface (i.e mirror)
i.e $\mathrm{r}=1$
i.e for this surface $100 \%$ of the photon will be reflected
i.e $p_f=-p_i$

So $|\Delta p|=\left|p_f-p_i\right|=\left|-p_i-p_i\right|=\frac{2 h}{\lambda}$
So Force is given as

$$
F=n(\Delta p)=\frac{P \lambda}{h c} * \frac{2 h}{\lambda}=\frac{2 P}{c}=\frac{2 I A}{c}
$$
 

and radiation pressure is given as

$$
\text { Pressure }=\frac{F}{A}=\frac{2 I}{c}
$$

- For neither perfectly reflecting nor perfectly absorbing body
i.e body having Absorption coefficient=a and reflection coefficient=r and we have $a+r=1$

So Force is given as

$$
\begin{aligned}
& F=\frac{a P}{c}+\frac{2 P r}{c}=\frac{P}{c}(a+2 r)=\frac{P}{c}((1-r)+2 r)=\frac{P}{c}(1+r) \\
& \text { Pressure }=\frac{F}{A}=\frac{P}{A c}(1+r)=\frac{I}{c}(1+r)
\end{aligned}
$$