Chapter 16.1, Problem 11SSC

To determine

The rate of lamp B if lamp A is rated 75 cd.

Answer to Problem 11SSC

The lamp $B$ is rated 27 cd.

Explanation of Solution

Given:

Two lamps are illuminating a screen equally.

Distance of lamp A from the screen is $d_A=$ 5.0 m.

Distance of the lamp B from the screen is $d_B=$ 3.0 m.

Rate of lamp $A$ is $75\text{cd}$ .

Formula used:

The expression for illumination is given by

  $E=\frac{P}{4\pi d^2}$  ...... (1)

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

Calculation:

Illuminance by the lamp $A$ is

  $E_A=\frac{P_A}{4\pi d_A{}^2}$  ...... (1)

Illuminance by the lamp $B$ is

  $E_B=\frac{P_B}{4\pi d_B{}^2}$  ...... (2)

Since both the lamp illuminate a screen equally so

  $E_A=E_A$  ...... (3)

Substitute $\frac{P_A}{4\pi d_A{}^2}$ for $E_A$ and $\frac{P_B}{4\pi d_B{}^2}$ for $E_B$ in equation (3)

  $\frac{P_A}{4\pi d_A{}^2}=\frac{P_B}{4\pi d_B{}^2}$  ...... (4)

The ratio $\frac{P}{4\pi }$ is the luminous intensity and is the rating of a lamp.

Therefore, rating of lamp $A$ and $B$ are $\frac{P_A}{4\pi }$ and $\frac{P_B}{4\pi }$ respectively.

Since $\frac{P_A}{4\pi }=75\text{cd}$

Substitute $26\text{cd}$ for $\frac{P_A}{4\pi }$ in equation (4)

  $\begin{array}{l}\frac{P_A}{4\pi d_A{}^2}=\frac{P_B}{4\pi d_B{}^2}\\ \frac{P_B}{4\pi }=d_B{}^2\left(\frac{P_A}{4\pi d_A{}^2}\right)\\ \frac{P_B}{4\pi }=3^2\left(\frac{75\text{cd}}{5^2}\right)\end{array}$

  $\boxed{\frac{P_B}{4\pi }=27\text{cd}}$

Conclusion:

Hence, the lamp $B$ is rated 27 cd.