Chapter 26, Problem 73P

(a)

To determine

Radius of the mirror.

Answer to Problem 73P

The radius should be greater than or equal to $\underset{\_}{0.334\text{m}}$.

Explanation of Solution

Write the equation for power

    $P=IA$        (I)

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

Write the equation for area.

    $A=\pi R_a{}^2$        (II)

Here $R_a$ is the radius of the circular area of the mirror.

Substitute (II) in (I) and rewrite in terms of $R_a$.

    $\begin{array}{c}P=I\pi R_a{}^2\\ R_a=\sqrt{\frac{I\pi }{P}}\end{array}$        (III)

Conclusion:

Substitute $350\text{W}$ for $P$ and $1000\text{W}/{\text{m}}^{\text{2}}$ for $I$ in (III)

    $\begin{array}{c}{R}_a=\sqrt{\frac{\left(1000\text{W}/{\text{m}}^{\text{2}}\right)\pi }{350\text{W}}}\\ =0.334\end{array}$

The radius should be greater than or equal to $\underset{\_}{0.334\text{m}}$.

(b)

To determine

Relationship between $R_a$ and $R$.

Answer to Problem 73P

The ratio of $R_a$ and $R$ is given by $\frac{R_a}{R}=\underset{\_}{0.0255}$.

Explanation of Solution

Write the mirror equation

  $\begin{array}{c}\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\ =\frac{2}{R}\end{array}$        (IV)

Here $f$ is the focal length, $R$ is the radius of curvature, $p$ is the object distance and $q$ is the image distance.

Write the equation for magnification

    $\begin{array}{c}M=\frac{h^{\prime }}{h}\\ =\frac{-q}{p}\end{array}$        (V)

Here $h^{\prime }$ is the image height, $h$ is the object height, $p$ is the object distance and $q$ is the image distance.

Object distance is infinite.

Then,

        $q=\frac{R}{2}$        (VI)

Rearrange (V) in terms of $h^{\prime }$ and substitute for $q$.

    $\begin{array}{c}{h}^{\prime }=-\frac{q}{p}h\\ =-\frac{R}{2}\left(\frac{h}{p}\right)\end{array}$        (VII)

Here $\frac{h}{p}$ is the angle that sun subtends. Substitute $0.533°$ for $\frac{h}{p}$ in (VII)

    $\begin{array}{c}{h}^{\prime }=-\frac{R}{2}\left(0.533°\left(\frac{\pi \text{rad}}{180°}\right)\right)\\ =-\frac{R}{2}\left(9.3\text{m}\text{rad}\right)\end{array}$        (VIII)

Write the equation for intensity of the image

    $I^{\prime }=\frac{P}{\pi r^2}$        (IX)

Substitute $r=\frac{h^{\prime }}{2}$ in (IX))

    $I^{\prime }=\frac{P}{\pi {\left(\frac{h^{\prime }}{2}\right)}^2}$        (X)

Substitute (III) and (VIII) in (X)

    $\begin{array}{c}{I}^{\prime }=\frac{4I\pi R_a{}^2}{\pi {\left(R/2\right)}^2{\left(9.3\times {10}^{-3}\text{rad}\right)}^2}\\ \frac{R_a}{R}=\sqrt{\frac{I^{\prime }{\left(9.3\times {10}^{-3}\text{rad}\right)}^2}{16I}}\end{array}$        (XI)

Conclusion:

Substitute  $1000\text{W}/{\text{m}}^{\text{2}}$ for $I$ and $120\times {10}^3\text{W}/{\text{m}}^{\text{2}}$ for $I^{\prime }$ in (XI)

    $\begin{array}{c}\frac{R_a}{R}=\sqrt{\frac{\left(120\times {10}^3\text{W}/{\text{m}}^{\text{2}}\right){\left(9.3\times {10}^{-3}\text{rad}\right)}^2}{16\left(1000\text{W}/{\text{m}}^{\text{2}}\right)}}\\ =0.0255\end{array}$

The ratio of $R_a$ and $R$ is given by $\frac{R_a}{R}=\underset{\_}{0.0255}$.