Chapter 16, Problem 59A

(a)

To determine

To identify:Whether the luminance at the book remains same.

Answer to Problem 59A

No, the luminance at the book will not be same. It reduces when the distance between the book and the lamp is doubled.

Explanation of Solution

Given:

The height of the reading lamp is $\text{35 cm}$ .

The distance between the book and the lamp is doubled.

Formula used:

The expression for the luminance $\left(E\right)$ of the lamp (point source) is,

  $E=\frac{P}{4\pi r^2}$

Here, $P$ is the luminous flux and $r$ is the distance between the lamp and the object.

Calculation:

The luminous flux of the lamp remains constant irrespective of the location of the book and lamp.The luminance produced by the lamp is inversely proportional to the distance between the lamp and the book, as other terms of the expression are constant.

When the distance between the lamp and the book is increased, the luminance of the lamp at the book decreases.

Hence, the luminance at the book reduces when the distance between the book and the lamp is doubled.

Conclusion:

Thus, the luminance at the book reduces when the distance between the book and the lamp is doubled.

(b)

To determine

The amount of increase or decrease in luminance at the book.

Answer to Problem 59A

The luminance at the book reduces to one-fourth when the distance between the book and the lamp is doubled.

Explanation of Solution

Given:

The height of the reading lamp is $\text{35 cm}$ .

The distance between the book and the lamp is doubled.

Formula:

The expression for the luminance $\left(E\right)$ of the lamp (point source) is,

  $E=\frac{P}{4\pi r^2}$

Here, $P$ is the luminous flux and $r$ is the distance between the lamp and the object.

Calculation:

Initially the distance between the lamp and the book is denoted by $r_{\text{1}}$ and the corresponding luminance is denoted by $E_1$ .

So, the luminance $\left(E_1\right)$ at the book is,

  $E_{\text{1}}=\frac{P}{4\pi r_1^2}$

Consider the distance between the lamp and the book is doubled, so the new distance between the lamp and the book is $r_{\text{2}}=2r_{\text{1}}$ and the corresponding luminance is denoted by $E_{\text{2}}$ .

The luminance $\left(E_2\right)$ at the book is,

  $\begin{array}{c}{E}_{\text{2}}=\frac{P}{4\pi r_2^2}\\ =\frac{P}{4\pi {\left(2r_1\right)}^2}\\ =\frac{1}{4}\times \left(\frac{P}{4\pi r_1^2}\right)\\ =\frac{1}{4}\times E_{\text{1}}\end{array}$

It is observed that the new luminance value is one-fourth of the old value.

Conclusion:

Thus, the luminance at the book reduces to one-fourth when the distance between the book and the lamp is doubled.