Physics Help Forum Inverse Square Law for a flat surface

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 Dec 18th 2013, 09:17 AM #1 Junior Member   Join Date: Dec 2013 Posts: 6 Inverse Square Law for a flat surface Hello This is my first post on this forum so my apologies ahead of time if I have posted it in the wrong place. I am currently looking into the inverse square law for a point source. It is Intensity is inversely proportional to the distance from the source squared. I then imagined the light landing on a flat plane. As the light hitting the nearest point to the plane will have traveled a shorter distance it will have a higher intensity than light hitting further along the plane. By making a right handed triangle (see attachment) between the point source, the plane and the point of incidence of the light we can see that the relationship between the minimum distance between the light source and the plane, R, and the distance between the light source and the point of incidence, D, is: D=R/cos(theta) Where theta is the the angle between the two displacements (see attachment). Adding this into the inverse square law suggests that for a point source shedding light onto a flat plane the intensity is proportional to cos(theta)/r all squared. Is this correct? I believed it was until I found a link that suggested that only the r was squared not the cosine. It is posted below. http://www.lepla.org/en/modules/Acti...7/m17-theo.htm I know the problem is probably very simple but any and all help would be greatly appreciated Thank you Attached Thumbnails
Dec 18th 2013, 09:56 AM   #2
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,344
 Originally Posted by Zook104 Is this correct?
Not quite - the amount of light energy that impinges on the plate per square meter must be multipled by the angle of incidence of the light onto the plate, per the article you cited. For a plate that is facing the source as in your example you get E = cos^3(theta)/r^2. However, note that for the case where the plate is small compared to r the max angle of theta is essentially zero, and so cos(theta)=1. The article assumes this is the case, so leaves out the complication of different parts of the plate being at different distances from the source, and only deals with the tilt of the plate with respect to the source.

 Dec 18th 2013, 11:59 AM #3 Junior Member   Join Date: Dec 2013 Posts: 6 Could you please run that by me again? I am having a little bit of a hard time understanding where the additional order of cos(theta) comes from. Many Thanks
 Dec 19th 2013, 04:54 AM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 It's based on the fact that for a plate tiltedf at angle phi away from the light source, the result is the amount of luminance is multiplied by cos(phi), as described in the web site you referenced. If you understand that part OK, then imagine you're an observer standing on a long flat plate, with a source of light directly above the center of the plate. If you stand at the center the source is directly overhead (at angle phi = 0). The amount of light energy per square inch that hits the plate at that point is inversely proportional to the distance of the source squared. Now imagine you move away from the center, walking towards one end, and get to a point where the source is no longer directly overhead but rather is at 30 degrees above the horizon - that is, the angle phi from the vertical s 60 degrees. Now you are at a distance from the source of Rcos^60, and since the light source is angled at 60 degrees it's intensity per unit area of the plate is cos(60) times what is was when the source was directly overhead. Thus the total amount of light power per unit area of the plate is proportional to cos(60) and inversely proportional to the distance squared, which as you pointed out is r/cos(60). Thus the power per unit area of the plate s proportional to cos^3(60)/r^2. I should point out that this is a calculation of power per unit area. To find the total amount of light power hitting the plate you would need to set up an integral in cos^3(theta) with theta varying from one end of the plate to the other.
 Dec 19th 2013, 06:05 AM #5 Junior Member   Join Date: Dec 2013 Posts: 6 Thank you so much I get it now

 Tags flat, inverse, law, square, surface