Question

About 6% of the power of a 100 W light bulb is converted to visible radiation. The average intensity of visible radiation at a distance of 8 m is (Assume that the radiation is emitted isotropically and neglect reflection

• A. $3.5\times {10}^{-3}W{m}^{-2}$
• B. $5.1\times {10}^{-3}W{m}^{-2}$
• C. $7.2\times {10}^{-3}W{m}^{-2}$
• D. $2.3\times {10}^{-3}W{m}^{-2}$

Answer 1

The correct option is C $7.2\times {10}^{-3}W{m}^{-2}$

Given: The power of the bulb is $100W$.

The amount of power converted to visible radiation is $6\%$.

To find the average intensity of visible radiations.

The amount of power obtained from visible light is:

$P^{\prime }=6\%of100W=\frac{6}{100}\times 100$

The intensity of the visible radiation at the distance of $8\text{m}$ is:

$I=\frac{Powerofvisiblelight}{area}$

$\Rightarrow \frac{100\times 6/100}{4\pi (8{)}^2}\Rightarrow 7.2\times {10}^{-3}W/m^2$

Option $\left(C\right)$ is correct.