91Ó°ÊÓ

Given that power P = 2.6 W and the diameter of its cylinder is d = 2.6 mm.. The intensity I is proportional to the area A and it represents the incident power P per area A.

${\mathrm{l}}_{\mathrm{A}}=\frac{\mathrm{P}}{\mathrm{A}}=\frac{\mathrm{P}}{4{\mathrm{Ï„Ï„°ù}}^2}$

The diameter of the laser is d =2.6 mm Hence, the radius is r = 1.3 mm and its area is calculated by

$\mathrm{A}={\mathrm{Ï€°ù}}^2=\mathrm{Ï€}(1.3×{10}^{-3}\mathrm{m}{)}^2=5.31×{10}^{-6}{\mathrm{m}}^2$

Substitute the values we have,

${\mathrm{l}}_{\mathrm{A}}=\frac{\mathrm{P}}{\mathrm{A}}=\frac{2.6\mathrm{W}}{5.31×{10}^{-6}\mathrm{m}}=48.97×{10}^4\mathrm{W}/{\mathrm{m}}^2$

Therefore, the intensity of Laser A is${\mathrm{I}}_{\mathrm{A}}=48.97×{10}^4\mathrm{W}/{\mathrm{m}}^2$

The intensity of a sinusoidal electromagnetic wave in a vacuum is related to the

electric-field amplitude E and it is given by equation${\mathrm{I}}_{\mathrm{B}}=\frac{1}{2}{\mathrm{Î\mu }}_0{{\mathrm{cE}}_{\max }}^2$Where${\mathrm{Î\mu }}_0$is the electric constant, c is the speed of light. The intensity is proportional to${\mathrm{E}}_{\max }^2$

Substitute the values we have,

$$\begin{gathered} {\mathrm{l}}_{\mathrm{B}}=\frac{1}{2}\left(8.854×{10}^{-12}\right)\left(3×{10}^8\right)\left(480\right) \\ =306\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the intensity of Laser B is$306\mathrm{W}/{\mathrm{m}}^2$

The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by equation in the form ${\mathrm{B}}_{\max }=\frac{{\mathrm{E}}_{\max }}{\mathrm{c}}$We know from the above equation that $\mathrm{l}=\frac{1}{2}{\mathrm{Î\mu }}_0{\mathrm{cE}}_{\max }^2=\frac{1}{2}{\mathrm{Î\mu }}_0\mathrm{c}{\left({\mathrm{cB}}_{\max }\right)}^2=\frac{1}{2}{\mathrm{Î\mu }}_0{\mathrm{c}}^3{\mathrm{B}}_{\max }^2$Substitute the values we have,

$$\begin{gathered} {\mathrm{l}}_{\mathrm{c}}=\frac{1}{2}\left(8.854×{10}^{-12}\right)\left(3×{10}^8\right)\left(8.7×{10}^{-6}\right) \\ =904\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the intensity of Laser C is$9047\mathrm{W}/{\mathrm{m}}^2$

Given that the diameter by d = 1.8 mm (r =0.9 mm), the force of radiation for totally reflected is $F_{rad}6×{10}^{-6}$To calculate the radiation pressure $p_{rad}$by

$$\begin{gathered} {\mathrm{p}}_{\mathrm{rad}}=\frac{{\mathrm{F}}_{\mathrm{rad}}}{\mathrm{A}} \\ =\frac{6×{10}^{-6}}{\mathrm{π}{\left(0.9×{10}^{-3}\right)}^2} \\ =0.0235\mathrm{Pa} \end{gathered}$$

The wave exerts an average force F per unit area and this is the radiation pressure Prad and it is the average value of dp/dt divided by the area. So, the radiation pressure of the wave that totally reflected is given by equation in the form$p_{rad}=\frac{2l_D}{c}$

Substitute the values we have

$$\begin{gathered} {\mathrm{l}}_{\mathrm{D}}=\frac{{\mathrm{cp}}_{\mathrm{rad}}}{2} \\ =\frac{\left(3×{10}^8\right)\left(0.0235\right)}{2}=3.5×{10}^6\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the intensity of Laser D is$3.5×{10}^{6}W/m^2$

The average energy density is given by u= 3 x 10¯7 J/m3. The intensity is related to $\mathrm{μ}$by$l_E=\mathrm{μ}c$Substitute the values we have,

$$\begin{gathered} {\mathrm{l}}_{\mathrm{E}}=\mathrm{쳦} \\ =3×{10}^{-7}\left(3×{10}^8\right)=90\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the intensity of Laser E=$90\mathrm{W}/{\mathrm{m}}^2$

From the results, the order of intensities of all lasers is$\mathrm{D}≻\mathrm{A}≻\mathrm{C}≻\mathrm{B}≻\mathrm{E}$