Physics > Basic Modern Physics > 8.0 Force exerted by a light beam on a surface
Basic Modern Physics
1.0 Photon theory of light
2.0 Characteristics of photon
3.0 Wave Particle Duality
4.0 Emission of electrons
5.0 Photoelectric Effect
5.1 Laws of Photoelectric emission
5.2 Photoelectric equation
5.3 Photoelectric Current
5.4 Stopping potential
5.5 Graph between $K{E_{max}}$ and frequency
6.0 Radiation Pressure And Force
7.0 Photon Density
8.0 Force exerted by a light beam on a surface
9.0 Early Atomic Structures
10.0 Bohr Model of The Hydrogen Atom
10.1 Radius of Orbit
10.2 Velocity of electron in the $n^th$ orbit
10.3 Orbital frequency of electron
11.0 Energy of electron in the $n^{th}$ orbit
12.0 Basic Definitions
13.0 Atomic Excitation
8.2 Surface exposed to radiation inclined at an angle
5.2 Photoelectric equation
5.3 Photoelectric Current
5.4 Stopping potential
5.5 Graph between $K{E_{max}}$ and frequency
10.2 Velocity of electron in the $n^th$ orbit
10.3 Orbital frequency of electron
Energy incident per unit time $ = IA\cos \theta $
$$\begin{equation} \begin{aligned} \therefore {\text{Number of photons emitted per unit time}}\quad N = \frac{{IA\lambda \cos \theta }}{{hc}} \\ \\\end{aligned} \end{equation} $$
| Case 1. The whole amount of radiation incident is absorbed by the surface. | Case 2. The whole amount of radiation incident is reflected by the surface. | Case 3. Some part of incident radiation is absorbed and rest is reflected by the surface. |
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So, $a = 1$ and $b = 0$. $$\begin{equation} \begin{aligned} F\sin \theta = \frac{{IA\cos \theta sin\theta }}{c} \\ \\\end{aligned} \end{equation} $$ | So, $a = 0$ and $b = 1$. Since, only the vertical component of velocity of the photon changes, momentum of the photon changes only in vertical direction and thus, the force is exerted on the plate only in vertical direction. Therefore, force on photons = Total change in momentum per unit time $$\begin{equation} \begin{aligned} = N\frac{h}{\lambda } = \frac{{IA\lambda \cos \theta }}{{hc}}\frac{{2h}}{\lambda }\cos \theta = \frac{{2IA{{\cos }^2}\theta }}{c} \\ \\\end{aligned} \end{equation} $$ Force on plate due to photons(from Newton Thirds Law), $$\begin{equation} \begin{aligned} F = \frac{{2IA{{\cos }^2}\theta }}{c} \\ \\\end{aligned} \end{equation} $$$$\begin{equation} \begin{aligned} {\text{Pressure}} = \frac{{2I{{\cos }^2}\theta }}{c} \\ \\\end{aligned} \end{equation} $$ | So, $0 < a < 1$ and $0 < r< 1$. Force on plate due to absorbed photons(from Newton Third Law), $$\begin{equation} \begin{aligned} F = \frac{{IA\cos \theta }}{c}(1 - r) \\ \\\end{aligned} \end{equation} $$ Force on plate due to reflected photons(from Newton Third Law), $$\begin{equation} \begin{aligned} F = \frac{{2IA{{\cos }^2}\theta }}{c}r \\ \\\end{aligned} \end{equation} $$ Component of force perpendicular to the surface ${F_1}$ $$\begin{equation} \begin{aligned} = \frac{{IA{{\cos }^2}\theta }}{c}2r + \frac{{2IA{{\cos }^2}\theta }}{c}(1 - r) \\ \\\end{aligned} \end{equation} $$Component of force parallel to the surface ${F_2}$ $$\begin{equation} \begin{aligned} = \frac{{IA\cos \theta sin\theta }}{c}(1 - r) \\ \\\end{aligned} \end{equation} $$ Resultant force on plate ${F_r}$= $\sqrt {{F_1}^2 + {F_2}^2} $ $$\begin{equation} \begin{aligned} = \frac{{IA\cos \theta }}{c}\sqrt {1 + {r^2} + 2r\cos \theta } \\ \\\end{aligned} \end{equation} $$$$\begin{equation} \begin{aligned} {\text{Pressure}} = \frac{{I{{\cos }^2}\theta }}{c}(1 + r) \\ \\\end{aligned} \end{equation} $$ |
Question 9. In the path of a uniform light beam of large cross sectional area and intensity $I$, a solid sphere of radius $R$ which is perfectly reflecting is placed. Find the force exerted on this sphere due to light beam.
Solution: The front surface of the sphere will be illuminated by the light as shown in $Fig .12$, as angle of incidence of light is different at different positions of the front face of the sphere.
We can consider a small elemental circular strip (ring) of angular width ${d\theta }$ on its surface at an angle $\theta $ from its horizontal diameter as shown. Let the area of this elemental strip is $dS$ then, $$dS = 2\theta R\,\,\sin \theta \,Rd\theta \quad ...(i)$$
where, $R\,\,\sin \theta $ is the radius of this circular strip (ring).
We take the projection of area of this slant elemental circular strip in vertical plane. Let $dA$ be the projection of the slant strip area $dS$ along the cross-sectional plane of the light beam and it is given as, $$dA = dS\,\cos \theta ...(ii)$$
Hence, the power of the light incident on this strip is $dP = IdA$. The momentum of photons per second incident on this strip is
$$dp = \frac{{dP}}{c} = \frac{{IdA}}{c}\quad ...(iii)$$
These photons are incident at an angle $\theta $ to the normal $N$ of this strip. As the surface of the sphere is perfectly reflecting, the photons will be reflected at the same angle $\theta $ to $N$ as shown in the figure 12.
Hence, the change in linear momentum of photons is along the normal and thus force exerted on this strip along the normal is
$$\begin{equation} \begin{aligned} dF = 2dp\,\cos \theta = \frac{{2IdA}}{c}\,\cos \theta ...(iv) \\ \\\end{aligned} \end{equation} $$
Thus, the net force on sphere will be along the $x$-axis only.
Thus, the force on the sphere will be given as
$$\begin{equation} \begin{aligned} F = \int {dF} \,\cos \theta = \int {\frac{{2IdA}}{c}} \,{\cos ^2}\theta \\ \quad = \int\limits_0^{\pi /2} {\frac{{2I}}{c}\left( {2\pi R\,\sin \theta \,\cos \theta \,Rd\theta } \right)} {\cos ^2}\theta \\ \quad = \frac{{4I\pi {R^2}}}{c}\int\limits_0^{\pi /2} {{{\cos }^3}\theta } \,\sin \theta \,d\theta = \frac{{4I\pi {R^2}}}{c}\left[ { - \frac{{{{\cos }^4}\theta }}{4}} \right]_0^{\pi /2} \\ \quad = \frac{{I\pi {R^2}}}{c}\left[ {1 - 0} \right] = \frac{{I\pi {R^2}}}{c} \\\end{aligned} \end{equation} $$
