Chapter 16, Problem 40A

To determine

To Find: The lightbulb’s luminous flux.

Answer to Problem 40A

The light bulb’s luminous flux is $1307 \mathrm{lm}$ .

Explanation of Solution

Given:

The minimum luminous flux is ${\text{P}}_{\text{1}}\text{=1750}\text{lm}$

Distance of the lamp above the paper is ${\text{r}}_1=1.25\text{m}$

Distance of the lightbulb from the paper is ${\text{r}}_2=1.08\text{m}$

Both the lamp and the light bulb illuminate the sheet of paper by the same amount.

Formula:

The expression for illumination is

  $\text{E=}\frac{\text{P}}{{\text{4πr}}^{\text{2}}}\text{lx}$  ...... (1)

Here r is the distance of the object from the source and P is the luminous flux.

Calculation:

Illumination caused by the lamp is equal = ${\text{E}}_1$ , Illumination cause by the light bulb = ${\text{E}}_2$

We also know that

  ${\text{E}}_1= {\text{E}}_2$

  $\frac{{\text{P}}_1}{{\text{4πr}}_1^2}=\frac{{\text{P}}_2}{{\text{4πr}}_2^2}$  ...... (2)

Here ${\text{P}}_1$ is the luminous flux of the lamp and ${\text{P}}_2$ is the luminous flux of the light.

Substitute $1750\text{lm}$ for ${\text{P}}_{\text{1}}$ , $1.25\text{m}$ for ${\text{r}}_1$ and $1.08\text{m}$ for ${\text{r}}_2$ in equation (2)

  $\begin{array}{c}\frac{1750}{4\times \text{π}\times {1.25}^2} =\frac{{\text{P}}_2}{4\times \text{π}\times {1.08}^2}\\ \text{}\text{}{\text{P}}_2=\frac{{\left(1.08 \text{m}\right)}^2\left(1750 \mathrm{lm}\right)}{{\left(1.25 \text{m}\right)}^2}\\ \text{}{\text{P}}_2=1307 \mathrm{lm}\end{array}$

Conclusion:

Hence, the required light bulb’s luminous flux is $1307 \mathrm{lm}$