Question

About $5\%$ of the power of a $100W$ light bulb is converted to visible radiation. What is the average intensity of visible radiation at distance of $10m$
A. $0.4\,W/{m^2}$
B. $0.04\,W/{m^2}$
C. $0.004\,W/{m^2}$
D. $0.0004\,W/{m^2}$

Answer 1

You can easily solve this question by considering the bulb as a point source and then applying the formula of intensity, since intensity is basically the distribution of power per unit area, the formula is $I\, = \,\dfrac{P}{A}$, where $I$ is the intensity, $P$ is the power and $A$ is the area it is covering at that distance.

Complete step-by-step solution:
We will be trying to solve this question exactly as we told in the hint section of the solution to this question. Firstly, we will consider the bulb as a point source of light, which will enable us to consider that the wavefronts are spherical wave fronts and will help us to reach the answer correctly and fast.
But first, we need to process the information that only $5\%$ of the total power is actually the power of visible radiation.
So, we can write it as:
Total power of the bulb, as given to us by the question, ${P_o} = 100W$
The power of visible radiation is: ${P_v}\, = \,\dfrac{5}{{100}} \times 100W\, = \,5W$
Now that we have the power of visible radiation, we can easily find the intensity of visible radiation at the given distance of $10m$ .
First, let’s discuss the formula of intensity:
$I\, = \,\dfrac{P}{A}$
We have already reached at the fact that the wavefronts are spherical, so that area will be given as:
$A = 4\pi {d^2}$, where $d$ is the distance from the point source, given as $10m$ in the question.
For our case, $P = {P_v} = 5W$
Now, all we have to do is to substitute the values in the formula:
$I\, = \,\dfrac{5}{{4\pi {{\left( {10} \right)}^2}}}\,W/{m^2}$
After simplifying or solving, we get:
$I = 0.00398\,W/{m^2} \\\ I \approx 0.004\,W/{m^2} \\\$
We can clearly see that the option C is the correct option as our answer matches it.

Note:- Many students don’t get the idea to consider the bulb as a simple pointed source of light and thus are unable to consider the wavefronts as spherical and thus reach at a totally different and wrong answer. Another important step in the solution is to know that the total power and power of visible radiation are two different things and we need either of them to reach the answer as asked in the question.