Chapter 26, Problem 59P

(a)

To determine

Intensity of light at the surface of the bulb.

Answer to Problem 59P

The intensity is $\underset{\_}{1.4\text{kW}/{\text{m}}^{\text{2}}}$.

Explanation of Solution

Write the equation for intensity

    $I=\frac{P}{A}$        (I)

Here $I$ is the intensity, $P$ is the power and $A$ is the area

Write the equation for surface area.

    $A=4\pi {\left(\frac{d}{2}\right)}^2$        (II)

Here $d$ is the diameter of the bulb.

Substitute (II) in (I)

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

Conclusion:

Substitute $4.5\text{W}$ for $P$, $3.2\text{cm}$ for $d$ in(III)

    $\begin{array}{c}I=\frac{4.5\text{W}}{\pi {\left(3.2\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\\ =\frac{4.5\text{W}}{\pi {\left(3.2\times {10}^{-2}\text{m}\right)}^2}\\ =1.4\text{kW}/{\text{m}}^{\text{2}}\end{array}$

The intensity is $\underset{\_}{1.4\text{kW}/{\text{m}}^{\text{2}}}$.

(b)

To determine

Light intensity at $7.2\text{cm}$ away from the center.

Answer to Problem 59P

The intensity is $\underset{\_}{6.91\text{mW}/{\text{m}}^{\text{2}}}$.

Explanation of Solution

Write the equation for radius

    $r=\frac{d}{2}$        (IV)

Here $d$ is the diameter and $r$ is the radius

Substitute (IV) in (III)

    $I=\frac{P}{4\pi r^2}$        (V)

Conclusion:

Substitute $4.5\text{W}$ for $P$, $7.2\text{m}$ for $r$ in (V)

    $\begin{array}{c}I=\frac{4.5\text{W}}{4\pi {\left(7.2\text{m}\right)}^2}\\ =6.91\text{mW}/{\text{m}}^{\text{2}}\end{array}$

The intensity is $\underset{\_}{6.91\text{mW}/{\text{m}}^{\text{2}}}$.

(c)

To determine

Diameter of the bulb’s image.

Answer to Problem 59P

Image diameter is $\underset{\_}{0.164\text{cm}}$.

Explanation of Solution

Write the mirror equation

      $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$        (VI)

Here $p$ is the object distance, $q$ is the image distance and $f$ is the focal length.

Rearrange (VI) in terms of $q$

    $q=\frac{pf}{p-f}$        (VII)

Diameter of the image is same as that of the height of the image.

Write the equation for image magnification

  $M=\frac{-q}{p}$        (VIII)

Here $M$ is the image magnification.

Write the equation for magnification in terms of image height.

    $M=\frac{h^{\prime }}{h}$        (IX)

Here $h\text{'}$ is the image height and $h$ is the object height.

Equate (VIII) and (IX) and write in terms of $h\text{'}$

    $h^{\prime }=-\frac{q}{p}h$        (X)

Conclusion:

Substitute $7.2\text{m}$ for $p$ and $0.35\text{m}$ for $f$ in (VII)

    $\begin{array}{c}q=\frac{\left(7.2\text{m}\right)\left(0.35\text{m}\right)}{7.2\text{m}-0.35\text{m}}\\ =0.368\text{m}\end{array}$

Substitute $0.389\text{m}$ for $q$, $7.2\text{m}$ for $p$ and $3.2\text{cm}$ for $h$ in (X)

    $\begin{array}{c}{h}^{\prime }=-\frac{0.389\text{m}}{7.2\text{m}}\left(3.2\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ =-\frac{0.389\text{m}}{7.2\text{m}}\left(3.2\times {10}^{-2}\text{m}\right)\\ =0.164\text{cm}\end{array}$

Image diameter is $\underset{\_}{0.164\text{cm}}$.

(d)

To determine

Light intensity at the image.

Answer to Problem 59P

The intensity is $\underset{\_}{58.1\text{W}/{\text{m}}^{\text{2}}}$.

Explanation of Solution

Write the equation for area of the bulb

    $A=\pi {\left(\frac{d}{2}\right)}^2$        (XI)

Here $d$ is the diameter.

Substitute (XI) in (I)

    $I=\frac{P}{\pi {\left(\frac{d}{2}\right)}^2}$        (XII)

Conclusion:

Substitute $6.91\text{mW}/{\text{m}}^{\text{2}}$ for $P$, $15\text{cm}$ for $d$ in (XII)

    $\begin{array}{c}I=\frac{6.91\text{mW}/{\text{m}}^{\text{2}}}{\pi {\left(\frac{15\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{2}\right)}^2}\\ =\frac{6.91\text{mW}/{\text{m}}^{\text{2}}}{\pi {\left(\frac{15\times {10}^{-2}\text{m}}{2}\right)}^2}\\ =58.1\text{W}/{\text{m}}^{\text{2}}\end{array}$

The intensity is $\underset{\_}{58.1\text{W}/{\text{m}}^{\text{2}}}$.