Chapter 16, Problem 57A

To determine

The comparison of the illuminance at screen B with the illuminance at screen A.

Answer to Problem 57A

  $E_B=\frac{E_A}{4}$

Explanation of Solution

Given:

The distance of screen A from the bulb, $r_A=2\text{}\text{m}$

The distance of screen B from the bulb, $r_B=4\text{}\text{m}$

Formula used:

The illuminance of a source at a distance $r$ is given by,

  $E=\frac{P}{4\pi r^2}$

Where $P=\text{luminous flux of the source}$

Calculation:

Now, since the bulb is same thus the luminous flux will remain same for both screen A and B.

Then, the illuminance at screen B will be,

  $E_B=\frac{P}{4\pi r_B^2}$

And, the illuminance at screen A will be,

  $E_A=\frac{P}{4\pi r_A^2}$

Then, dividing above two equations,

  $\begin{array}{l}\frac{E_B}{E_A}=\frac{\left(\frac{P}{4\pi r_B^2}\right)}{\left(\frac{P}{4\pi r_A^2}\right)}\\ \frac{E_B}{E_A}=\frac{r_A^2}{r_B^2}\end{array}$

Using given values in above,

  $\begin{array}{c}\frac{E_B}{E_A}=\frac{r_A^2}{r_B^2}\\ \frac{E_B}{E_A}=\frac{{\left(2\right)}^2}{{\left(4\right)}^2}\\ E_B=\frac{E_A}{4}\end{array}$

Conclusion:

Hence, the illuminance at screen B is one-fourth times of the illuminance at screen A.