Chapter 22, Problem 70P

(a)

To determine

The intensity of laser beam.

Answer to Problem 70P

Intensity is $1.1\text{kW}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Write the equation for intensity of laser beam.

    $I=\frac{P}{A}$

Here, $I$ is the intensity of laser beam, $P$ is the power, and $A$ is the area of cross section of beam.

Write the equation to find $A$.

    $A=\frac{1}{4}\pi d^2$

Here, $d$ is the diameter of beam.

Rewrite the equation for $I$ by substituting the above relation for $A$.

  $\begin{array}{c}I=\frac{P}{\frac{1}{4}\pi d^2}\\ =\frac{4P}{\pi d^2}\end{array}$                                                                                                                 (I)

Conclusion:

Substitute $2.0\text{mW}$ for $P$, $3.14$ for $\pi$ and $1.5\text{mm}$ for $d$ in the above equation to find $I$.

  $\begin{array}{c}I=\frac{4\left(2.0\text{mW}\left(\frac{{10}^{-3}\text{W}}{1\text{mW}}\right)\right)}{\left(3.14\right){\left(1.5\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^2}\\ =\frac{8.0\times {10}^{-3}\text{W}}{7.065\times {10}^{-3}\text{m}}\\ =1100\text{W}/{\text{m}}^{\text{2}}\left(\frac{\text{1}\text{kW}}{{\text{10}}^{\text{3}}{\text{m}}^{\text{2}}}\right)\\ =1.1\text{kW}/{\text{m}}^{\text{2}}\end{array}$

Therefore, the rms current is $20\text{A}$.

(b)

To determine

The intensity of light incident on the retina.

Answer to Problem 70P

Intensity is $6.4\text{MW}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Conclusion:

Substitute $2.0\text{mW}$ for $P$, $3.14$ for $\pi$ and $20.0\text{μm}$ for $d$ in equation (I) to find $I$.

  $\begin{array}{c}I=\frac{4\left(2.0\text{mW}\left(\frac{{10}^{-3}\text{W}}{1\text{mW}}\right)\right)}{\left(3.14\right){\left(20.0\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right)}^2}\\ =\frac{8.0\times {10}^{-3}\text{W}}{1.256\times {10}^{-9}\text{m}}\\ =6.4\times {10}^6\text{W}/{\text{m}}^{\text{2}}\left(\frac{\text{1}\text{MW}}{{\text{10}}^{\text{6}}\text{W}}\right)\\ =6.4\text{MW}/{\text{m}}^{\text{2}}\end{array}$

Therefore, the rms current is $20\text{A}$.

(c)

To determine

Total energy incident on the retina.

Answer to Problem 70P

Total energy is $0.16\text{mJ}$.

Explanation of Solution

Write the equation for total energy incident on the retina.

  $U=P\Delta t$

Here, $U$ is the total energy incident on the retina, $P$ is the power, and $\Delta t$ is the time duration.

Conclusion:

Substitute $2.0\text{mW}$ for $P$ and $80\text{ms}$ for $\Delta t$ in the above equation to find $U$.

  $\begin{array}{c}U=\left(2.0\text{mW}\left(\frac{{10}^{-3}\text{W}}{1\text{mW}}\right)\right)\left(80\text{ms}\left(\frac{{10}^{-3}\text{s}}{1\text{ms}}\right)\right)\\ =160\times {10}^{-6}\text{J}\left(\frac{1\text{mJ}}{{10}^{-3}\text{J}}\right)\\ =0.16\text{mJ}\end{array}$

Therefore, the total energy is $0.16\text{mJ}$.