A laser beam of intensity I reflects from a flat, totally reflecting surface of area A, with a normal at angle $\mathit{\theta }$with the beam.Write an expression for the beam’s radiationpressure${\mathit{p}}_{\mathbf{r}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}$onthe surface in terms of the beam’s pressurewhen$\mathit{\theta }\mathbf{=}{\mathbf{0}}^{\mathbf{0}}$.

• Intensity of laser beam is$I$.
• Reflecting surface area is.$A$
• The angle between beam and incident beam is.$\theta$

By using the concept of the radiation pressure and the ray diagram for the reflection we can find the required expression.

Formulae:

Radiation pressure for the total reflection with the normal incident ray,

$p_r=\frac{F_N}{A}=\frac{2I_N}{C}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

The trigonometric formula of cosine angle,$\cos \theta =\frac{\text{adjacentside}}{\text{Hypotenuse}}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

When the incident ray makes an angle with the surface, then the situation will be as shown.

In$\Delta ABO$, the angle of$\Delta ABO$is given using equation (2) as follows:

$\begin{array}{c}\cos \theta =\frac{F_N}{F_I}\\ F_N=\frac{F_I}{\cos \theta }\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

In,$\Delta BCO$the angle of $\Delta BCO$is given using equation (2) as follows:

$\begin{array}{c}\cos \theta =\frac{I_N}{I}\\ I_N=I\cos \theta \cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)\end{array}$

Using equations (3) and (4) in equation (1), we can get that

Hence, the expression of radiation of pressure is.$p_r\left(\theta \right)=p_{r\perp }\left({\cos }^2\theta \right)$