Recent content by digitaljeff

argh forgot about the whole "virtual" image thing.. virtual image to screen would be 3 units ... as source to mirror is 1 unit.. then virtual image to mirror another 1 unit.. then 1 unit to screen.. sooo.. 3 units from virtual image to screen... 1 unit from source to screen..?

image to source would be 2 units, so intensity would be decreased by a factor of 4 for the light coming from the image on the screen compared to light coming from the source on the screen? .. argh sorry if I am clueless..

decreases by the square of the distance from the source?? i think.. just re-reading my textbook trying to figure it out

i would think intensity in this case would be power/distance, so S/d originally. then with the mirror S/d + S/2d ... am i on the right track??

Homework Statement You put a point source of light (S) a distance (d) in front of screen (A). How is the light intensity at the center of the screen changed if you put a completely reflecting mirror (M) a distance (d) behind the source? M-----d-----S-----d-----A Homework Equations...