Chapter 36, Problem 36.83AP

(a)

To determine

The light intensity at the surface of the light bulb.

Answer to Problem 36.83AP

The light intensity at the surface of the light bulb is $1.40\text{kW}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given info:

The diameter of the spherical light bulb is $3.20\text{cm}$ an the power of light bulb is $4.50\text{W}$.

Formula to calculate the intensity at the surface of the light bulb is,

$I=\frac{P}{4\pi {\left(\frac{d}{2}\right)}^2}$

Here,

$P$ is the power of light bulb.

$d$ is the diameter of the spherical light bulb.

Substitute $3.20\text{cm}$ for $d$ and $4.50\text{W}$ for $P$ in the above equation.

$\begin{array}{c}I=\frac{4.50\text{W}}{4\pi {\left(\frac{3.20\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{2}\right)}^2}\\ =1398.8\text{W}/{\text{m}}^{\text{2}}\left(\frac{1\text{kW}/{\text{m}}^{\text{2}}}{{10}^{-3}\text{W}/{\text{m}}^{\text{2}}}\right)\\ \approx 1.40\text{kW}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the light intensity at the surface of the light bulb is $1.40\text{kW}/{\text{m}}^{\text{2}}$.

(b)

To determine

The light intensity at a distance $7.20\text{m}$ from the center of the light bulb.

Answer to Problem 36.83AP

The light intensity at a distance $7.20\text{m}$ from the center of the light bulb is $6.91\text{mW}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given info: The diameter of the spherical light bulb is $3.20\text{cm}$ an the power of light bulb is $4.50\text{W}$.

Formula to calculate the intensity at a distance $7.20\text{m}$ from the center of the light bulb,

,

$I^{\prime }=\frac{P}{4\pi {\left(r\right)}^2}$

Here,

$P$ is the power of light bulb.

$r$ is the distance from the center of the light bulb.

Substitute $7.20\text{m}$ for $r$ and $4.50\text{W}$ for $P$ in the above equation.

$\begin{array}{c}{I}^{\prime }=\frac{4.50\text{W}}{4\pi {\left(7.20\text{m}\right)}^2}\\ =6.91\text{mW}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the light intensity at a distance $7.20\text{m}$ from the center of the light bulb is $6.91\text{mW}/{\text{m}}^{\text{2}}$.

(c)

To determine

The diameter of the light bulb image.

Answer to Problem 36.83AP

The diameter of the light bulb image is $0.164\text{cm}$.

Explanation of Solution

Given info: The diameter of the spherical light bulb is $3.20\text{cm}$,  the power of light bulb is $4.50\text{W}$, the diameter of the circular lens is $15.0\text{cm}$, the focal length of the lens is $35.0\text{cm}$ and the object distance is $7.2\text{m}$.

From the lens maker formula,

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

Here,

$p$ is the object distance.

$q$ is the image distance.

$f$ is the focal length.

Substitute $7.2\text{m}$ for $p$ and $35.0\text{cm}$ for $f$ in the above equation.

$\begin{array}{c}\frac{1}{7.2\text{m}}+\frac{1}{q}=\frac{1}{35.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ q=0.368\text{m}\end{array}$

Thus, the image distance is $0.368\text{m}$.

Formula to calculate the magnification of the image,

$M=\frac{r^{\prime }}{r}$

Here,

$r^{\prime }$ is the diameter of image of the spherical light bulb.

$r$ is the diameter of spherical light bulb.

Formula to calculate the magnification of the image in terms of image and object distance is,

$M=-\frac{q}{p}$

Here,

$q$ is the image distance.

$p$ is the object distance.

Equate the equation (1) and (2) and then rearrange for $r^{\prime }$,

$\begin{array}{l}\frac{r^{\prime }}{r}=-\frac{q}{p}\\ r^{\prime }=r\left(-\frac{q}{p}\right)\end{array}$ (3)

Substitute $0.368\text{m}$ for $q$, $7.2\text{m}$ for $p$ and $3.20\text{cm}$ for $r$ in the equation (3).

$\begin{array}{c}{r}^{\prime }=\left(3.20\text{cm}\right)\left(-\frac{0.368\text{m}}{7.2\text{m}}\right)\\ =-0.164\text{cm}\end{array}$

Conclusion:

Therefore, the diameter of the light bulb image is $0.164\text{cm}$.

(d)

To determine

The light intensity of the image.

Answer to Problem 36.83AP

The light intensity of the image is $58.1\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given info: The diameter of the spherical light bulb is $3.20\text{cm}$,  the power of light bulb is $4.50\text{W}$, the diameter of the circular lens is $15.0\text{cm}$, the focal length of the lens is $35.0\text{cm}$ and the object distance is $7.2\text{m}$.

From part (b) the light intensity at a distance $7.20\text{m}$ from the center of the light bulb is $6.91\text{mW}/{\text{m}}^{\text{2}}$.

From part (c) the diameter of the light bulb image is $0.164\text{cm}$.

Formula to calculate the power of lens is,

$P^{\prime }=I^{\prime }A$ (4)

Here,

$I^{\prime }$ is the intensity at a distance $7.20\text{m}$ from the center of the light bulb.

$A$ is the area of lens.

The area of the lens is,

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

Substitute $15.0\text{cm}$ for $d$ in the above equation.

$\begin{array}{c}A=\frac{\pi }{4}{\left[15.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2\\ =0.0176{\text{m}}^{\text{2}}\end{array}$

The area of the lens is $0.0176{\text{m}}^{\text{2}}$.

Substitute $6.91\text{mW}/{\text{m}}^{\text{2}}$ for $I^{\prime }$ and $0.0176{\text{m}}^{\text{2}}$ for $A$ in the equation (4).

$\begin{array}{c}{P}^{\prime }=\left(6.91\times {\text{10}}^{-3}\text{W}/{\text{m}}^{\text{2}}\right)\left(0.0176{\text{m}}^{\text{2}}\right)\\ =0.121\times {\text{10}}^{-3}\text{W}\end{array}$

Formula to calculate the required intensity of bulb light image,

$I^{″}=\frac{P^{\prime }}{A_{\text{image}}}$ (5)

The area of image is,

$A_{\text{image}}=\frac{\pi }{4}{\left(h^{\prime }\right)}^2$

Substitute $0.164\text{cm}$ for $h^{\prime }$ in the above equation.

$\begin{array}{c}{A}_{\text{image}}=\frac{\pi }{4}{\left[0.164\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2\\ =0.021\times {10}^{-4}{\text{m}}^2\end{array}$

Substitute $0.021\times {10}^{-4}{\text{m}}^2$ for $A_{\text{image}}$ and $0.121\times {\text{10}}^{-3}\text{W}$ in the equation (5).

$\begin{array}{c}{I}^{″}=\frac{0.121\times {\text{10}}^{-3}\text{W}}{0.021\times {10}^{-4}{\text{m}}^2}\\ =58.1\text{W}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the light intensity of the image is $58.1\text{W}/{\text{m}}^{\text{2}}$.