Chapter 1, Problem 135RE

a.

To determine

To calculate:For the given information develop equation which relates intensity of illumination $I$ inversely to square of distance $d$

Answer to Problem 135RE

The required equation is $I=\frac{k}{d^2}$

Explanation of Solution

Given information:

The Intensity of illumination $I$ from a given light is inversely proportional to square of distance $d$ from the given light.

Formula used:

For 2 variables say, $x$ and $y$ , the statement $x$ is directly proportional to $y$ can be written as:

  $x\alpha y$

Which can be written as:

  $x=ky$

Where $k$ denotes the proportionality constant.

Similarly the statement $x$ is inversely proportional to $y$ can be interpreted as:

  $x\alpha \frac{1}{y}$

Which can be written as:

  $x=k\left(\frac{1}{y}\right)$

Where $k$ denotes the proportionality constant.

Calculation:

AsIntensity of illumination $I$ from a given light is inversely proportional to square of distance $d$ from the given light.

Recall, For 2 variables say, $x$ and $y$ , the statement $x$ is inversely proportional to $y$ can be interpreted as:

  $x\alpha \frac{1}{y}$

Which can be written as:

  $x=k\left(\frac{1}{y}\right)$

Where $k$ denotes the proportionality constant.

Hence, this variation can be expressed as follows:

  $\begin{array}{l}I\alpha \frac{1}{d^2}\\ I=\frac{k}{d^2}\end{array}$

Where $k$ denotes the proportionality constant.

Hence, the required equation is $I=\frac{k}{d^2}$

b.

To determine

To calculate:The constant of proportionality $k$ if the lamp intensity is $1000$ candles and the distance is $8$ meters.

Answer to Problem 135RE

The constant of proportionality $k$ if the lamp intensity is $1000$ candles and the distance is $8$ meters is $k=64000$ candles per $m^2$

Explanation of Solution

Given information:

The equation found is $I=\frac{k}{d^2}$ and the lamp intensity is $1000$ candles while the distance from the light is $8$ meters

Formula used:

For 2 variables say, $x$ and $y$ , the statement $x$ is directly proportional to $y$ can be written as:

  $x\alpha y$

Which can be written as:

  $x=ky$

Where $k$ denotes the proportionality constant.

Similarly the statement $x$ is inversely proportional to $y$ can be interpreted as:

  $x\alpha \frac{1}{y}$

Which can be written as:

  $x=k\left(\frac{1}{y}\right)$

Where $k$ denotes the proportionality constant.

Calculation:

AsIntensity of illumination $I$ from a given light is inversely proportional to square of distance $d$ from the given light.

Form a. the required equation is $I=\frac{k}{d^2}$ $\left(1\right)$

It is given that the intensity of illumination $I=1000$ and distance from the light $d=8$ meters.

Put these values in $\left(1\right)$ to get:

  $\begin{array}{l}1000=\frac{k}{8^2}\\ k=64*1000\\ k=64000\end{array}$

Hence, the constant of proportionality is $k=64000$ candles per $m^2$

c.

To determine

To calculate:The intensity of lamp when the distance from the light is $20$ meters

Answer to Problem 135RE

The intensity of lamp when the distance from the light is $20$ meters is $I=160$ candles.

Explanation of Solution

Given information:

The equation found is $I=\frac{k}{d^2}$ and the constant of proportionality is $k=64000$ candles per $m^2$ . Also distance from the light is $d=20$ meters

Formula used:

For 2 variables say, $x$ and $y$ , the statement $x$ is directly proportional to $y$ can be written as:

  $x\alpha y$

Which can be written as:

  $x=ky$

Where $k$ denotes the proportionality constant.

Similarly the statement $x$ is inversely proportional to $y$ can be interpreted as:

  $x\alpha \frac{1}{y}$

Which can be written as:

  $x=k\left(\frac{1}{y}\right)$

Where $k$ denotes the proportionality constant.

Calculation:

AsIntensity of illumination $I$ from a given light is inversely proportional to square of distance $d$ from the given light.

Form a. the required equation is $I=\frac{k}{d^2}$ $\left(1\right)$

From b. the constant of proportionality is $k=64000$ candles per $m^2$ $\left(2\right)$

Also, it is given distance from the light is $d=20$ meters $\left(3\right)$

Put $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$ to get:

  $\begin{array}{l}I=\frac{64000}{{20}^2}\\ I=\frac{64000}{400}\\ I=160\end{array}$

Hence, the intensity of lamp when the distance from the light is $20$ meters is $I=160$ candles.