Example 1: Calculating intensity and power: How much energy is in a ray of Sunlight?

The average intensity of sunlight on Earth’s surface is about[latex]\boldsymbol{700\textbf{ W/m}^2}.[/latex]

(a) Calculate the amount of energy that falls on a solar collector having an area of[latex]\boldsymbol{0.500\textbf{ m}^2}[/latex]in[latex]\boldsymbol{4.00\textbf{ h}}.[/latex]

(b) What intensity would such sunlight have if concentrated by a magnifying glass onto an area 200 times smaller than its own?

Strategy a

Because power is energy per unit time or[latex]\boldsymbol{P=\frac{E}{t}},[/latex]the definition of intensity can be written as[latex]\boldsymbol{I=\frac{P}{A}=\frac{E/t}{A}},[/latex]and this equation can be solved for E with the given information.

Solution a

1. Begin with the equation that states the definition of intensity:
   [latex]\boldsymbol{I\:=}[/latex][latex]\boldsymbol{\frac{P}{A}}.[/latex]
2. Replace[latex]\boldsymbol{P}[/latex]with its equivalent[latex]\boldsymbol{E/t}:[/latex]
   [latex]\boldsymbol{I\:=}[/latex][latex]\boldsymbol{\frac{E/t}{A}}.[/latex]
3. Solve for[latex]\boldsymbol{E}:[/latex]
   [latex]\boldsymbol{E=IAt}.[/latex]
4. Substitute known values into the equation:
   [latex]\boldsymbol{E=(700\textbf{ W/m}^2)(0.500\textbf{ m}^2)[(4.00\textbf{ h})(3600\textbf{ s/h})]}.[/latex]
5. Calculate to find[latex]\boldsymbol{E}[/latex]and convert units:
   [latex]\boldsymbol{5.04\times10^6\textbf{ J}},[/latex]

Discussion a

The energy falling on the solar collector in 4 h in part is enough to be useful—for example, for heating a significant amount of water.

Strategy b

Taking a ratio of new intensity to old intensity and using primes for the new quantities, we will find that it depends on the ratio of the areas. All other quantities will cancel.

Solution b

1. Take the ratio of intensities, which yields:
   [latex]\boldsymbol{\frac{I^{\prime}}{I}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{P^{\prime}/A^{\prime}}{P/A}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{A}{A^{\prime}}}([/latex][latex]\boldsymbol{\textbf{The powers cancel because }P^{\prime}=P}[/latex][latex]).[/latex]
2. Identify the knowns:
   [latex]\boldsymbol{A=200A^{\prime}},[/latex]
   [latex]\boldsymbol{\frac{I^{\prime}}{I}}[/latex][latex]\boldsymbol{=200}.[/latex]
3. Substitute known quantities:
   [latex]\boldsymbol{I^{\prime}=200I=200(700\textbf{ W/m}^2)}.[/latex]
4. Calculate to find[latex]\boldsymbol{I^{\prime}}:[/latex]

   [latex]\boldsymbol{I^{\prime}=1.40\times10^5\textbf{ W/m}^2}.[/latex]

Discussion b

Decreasing the area increases the intensity considerably. The intensity of the concentrated sunlight could even start a fire.

Example 2: Determine the combined intensity of two waves: Perfect constructive interference

If two identical waves, each having an intensity of[latex]\boldsymbol{1.00\textbf{ W/m}^2},[/latex]interfere perfectly constructively, what is the intensity of the resulting wave?

Strategy

We know from Chapter 16.10 Superposition and Interference that when two identical waves, which have equal amplitudes[latex]\boldsymbol{X},[/latex]interfere perfectly constructively, the resulting wave has an amplitude of[latex]\boldsymbol{2X}.[/latex]Because a wave’s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves.

Solution

1. Recall that intensity is proportional to amplitude squared.
2. Calculate the new amplitude:
   [latex]\boldsymbol{I^{\prime}\propto(X^{\prime})^2=(2X)^2=4X^2}.[/latex]
3. Recall that the intensity of the old amplitude was:
   [latex]\boldsymbol{I\propto{X}^2}.[/latex]
4. Take the ratio of new intensity to the old intensity. This gives:
   [latex]\boldsymbol{\frac{I^{\prime}}{I}}[/latex][latex]\boldsymbol{=4}.[/latex]
5. Calculate to find[latex]\boldsymbol{I^{\prime}}:[/latex]
   [latex]\boldsymbol{I^{\prime}=4I=4.00\textbf{ W/m}^2}.[/latex]

Discussion

The intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The two individual waves each have intensities of[latex]\boldsymbol{1.00\textbf{ W/m}^2},[/latex]yet their sum has an intensity of[latex]\boldsymbol{4.00\textbf{ W/m}^2},[/latex]which may appear to violate conservation of energy. This violation, of course, cannot happen. What does happen is intriguing. The area over which the intensity is[latex]\boldsymbol{4.00\textbf{ W/m}^2}[/latex]is much less than the area covered by the two waves before they interfered. There are other areas where the intensity is zero. The addition of waves is not as simple as our first look in Chapter 16.10 Superposition and Interference suggested. We actually get a pattern of both constructive interference and destructive interference whenever two waves are added. For example, if we have two stereo speakers putting out[latex]\boldsymbol{1.00\textbf{ W/m}^2}[/latex]each, there will be places in the room where the intensity is[latex]\boldsymbol{4.00\textbf{ W/m}^2},[/latex]other places where the intensity is zero, and others in between. Figure 2 shows what this interference might look like. We will pursue interference patterns elsewhere in this text.