Problem involving light as a particle

[lank20125]

I am having difficulties with a problem that asks for the number of photons per second that are absorbed by a piece of paper. I am given the power of the light source, the distance the source is away from the paper, the wavelength of the light, and the area of the piece of paper. I know that the number of photons per second absorbed can be found by dividing the power by energy of the photons. However, I am not sure how to relate the power of the light source to the distance it is away from the piece of paper and its area. Any help would be greatly appreciated. Thanks a lot!

 

[notsureanymore]

unless the light source diverges to the point where some of the light misses the paper wouldn't the distance be irelevant.

I assume there is no intervening matter to absorb the light.

 

[Schneibster]

For a first approximation, you need the total area of the sphere that is at that distance from the light; then you get the ratio of the area of the paper to the area of the sphere. This ratio will be less than one.

The power needs to be turned into energy; but since the power is almost certainly in watts, and you need the energy in a second, you can convert directly to joules from watts; joules are watt-seconds. So you have that right. You can then use Planck's constant to find the total number of photons, and multiply by the ratio of the areas.

The first approximation assumes that the paper is a section of a sphere, i.e. curved; if it is held flat, then the power it receives will be slightly less. If you have the dimensions of the piece of paper, rather than just the area, then you can use them to calculate the solid angle of the paper with trig. Its just the horizontal angle times the vertical angle; I suggest radians, the calculations will be a little easier. Once you have the solid angle, you can use the solid angle of a complete spherical field (you get to go look that up) and derive the ratio that way. It will be the most accurate answer; but be sure to give both approximations and explain your reasoning, because the instructor may not have anticipated this answer (although that is pretty unlikely- still, hedge your bets).

If you do not have the dimensions, the instructor probably did not intend you to use the second way. In this case, you could use calculus over the trig functions to find the curve that describes the solid angle for various ratios of width to height for the paper to the limits (don't forget- the paper cannot be greater in a single dimension than the diameter of a circle at the stated distance!), and probably turn in the chart of that curve and your interpolated points for the curve along with the first approximation answer for extra credit. But I really don't want to go dig out my calculus book to tell you how to do that calculation; so unless you've had some advanced calculus, you're on your own.