Chapter 16.1, Problem 1PP

To determine

To Compare: The illumination on the book before and after the lamp is moved.

Answer to Problem 1PP

Illumination after moving the lamp is one ninth of the original illumination i.e. $\boxed{E_{Before}=9\times E_{After}}$ .

Explanation of Solution

Given:

The lamp is moved from $30\text{cm}$ to $90\text{cm}$ .

Formula used:

The expression for illumination is given by

  $E=\frac{P}{4\pi r^2}$........(1)

Here, $P$ and $r$ are luminous flux and radius of the sphere which is equal to the distance of the object from the light source respectively.

Calculation:

Luminous flux for different distances will remain the same.

Before light is moved, the illumination can be calculated by using equation (1)

Substitute $30\text{cm}$ for $r$ in equation (1),

  $E_{before}=\frac{P}{4\pi {\left(30\text{cm}\right)}^2}$

  $E_{before}=\frac{P}{4\pi {\left(30\times {10}^{-2}\text{m}\right)}^2}$

  $E_{before}=\frac{\left(\frac{P}{4\pi }\right)}{{\left(0.30\text{m}\right)}^2}$........(2)

After light is moved, the illumination can be calculated by using equation (1)

Substitute $90\text{cm}$ for $r$ in equation (1),

  $E_{before}=\frac{P}{4\pi {\left(90\text{cm}\right)}^2}$

  $E_{before}=\frac{P}{4\pi {\left(60\times {10}^{-2}\text{m}\right)}^2}$

  $E_{after}=\frac{\left(\frac{P}{4\pi }\right)}{{\left(0.90\text{m}\right)}^2}$........(3)

Divide both sides of equation (3) by equation (2),

  $\begin{array}{l}\frac{E_{After}}{E_{Before}}=\frac{\cancel{\left(\frac{P}{4\pi }\right)}}{{\left(0.90\text{m}\right)}^2}\times \frac{{\left(0.30\text{m}\right)}^2}{\cancel{\left(\frac{P}{4\pi }\right)}}\\ \frac{E_{After}}{E_{Before}}=\frac{{\left(0.30\text{m}\right)}^2}{{\left(0.90\text{m}\right)}^2}\\ \frac{E_{After}}{E_{Before}}=\frac{0.09{\text{m}}^{\text{2}}}{0.81{\text{m}}^{\text{2}}}\end{array}$

  $\frac{E_{After}}{E_{Before}}=\frac{1}{9}$

Or $\boxed{E_{Before}=9\times E_{After}}$

Conclusion:

Hence, illumination after moving the lamp is one ninth of the original illumination i.e. $\boxed{E_{Before}=9\times E_{After}}$ .