Chapter 16.1, Problem 4PP

To determine

The distance of the second lamp from the screen.

Answer to Problem 4PP

The distance of the lamp 2 from the source is 3.2 m.

Explanation of Solution

Given:

Luminous flux of the first lamp is $1445\text{lm}$ .

The distance from of the screen of the first lamp is 2.5 m.

Luminous flux of the second lamp is $2375\text{lm}$ .

Formula used:

The expression for illumination is given by

  $E=\frac{P}{4\pi r^2}$

Here, $P$ and $r$ are luminous flux and the distance of the object from the light source respectively.

Calculation:

The illuminance of the lamp 1 is

  $E_1=\frac{P_1}{4\pi r_1{}^2}$  ...... (1)

Here, $r_1$ is the radius which is the distance of the lamp 1 from the screen.

The illuminance of the lamp 2 is

  $E_2=\frac{P_1}{4\pi r_2{}^2}$  ...... (2)

Here, $r_2$ is the radius which is the distance of the lamp 1 from the screen.

Since the illuminance of both the lamp is same i.e.

  $E_1=E_2$  ...... (3)

Substitute $\frac{P_1}{4\pi r_1{}^2}$ for $E_1$ and $\frac{P_1}{4\pi r_2{}^2}$ for $E_2$ in equation (3)

  $\begin{array}{l}\frac{P_1}{4\pi r_1{}^2}=\frac{P_1}{4\pi r_2{}^2}\\ \frac{P_1}{r_1{}^2}=\frac{P_1}{r_2{}^2}\end{array}$

  $r_2=\sqrt{\frac{P_2{r}_1{}^2}{P_1}}$  ...... (4)

Substitute 1445 lm for $P_1$ , 2375 lm for $P_2$ and 2.5 m for $r_1$ in equation (4)

  $\begin{array}{l}{r}_2=\sqrt{\frac{\left(2375\text{lm}\right)\times {\left(2.5\text{lm}\right)}^2}{\left(1445\text{lm}\right)}}\\ \boxed{r_2=3.2\text{m}}\end{array}$

Conclusion:

Hence, the distance of the lamp 2 from the source is 3.2 m.