Determining light intensity with a mirror

[digitaljeff]

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

I=Power/Area ??

The Attempt at a Solution

The answer it is giving in the book is 10/9 of the original intensity but i have no clue at all how to get this.

 

[kuruman]

When you put the mirror in place, the image constitutes a second source in addition to the original one. How far from the screen is this second source? How does intensity depend on distance?

 

[digitaljeff]

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??

 

[kuruman]

No. Answer me this, if you double the distance between the source and the screen, by what factor is the intensity on the screen reduced?

 

[digitaljeff]

No. Answer me this, if you double the distance between the source and the screen, by what factor is the intensity on the screen reduced?

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

 

[kuruman]

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

Correct. Now call the original screen-to-source distance one "unit". How many "units" is the image-to-source distance?

 

[digitaljeff]

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

 

[kuruman]

Two units is the mirror-to-source distance. The image is behind the mirror by an additional how many units?

 

[digitaljeff]

Two units is the mirror-to-source distance. The image is behind the mirror by an additional how many units?

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..?

 

[kuruman]

Correct. So if the intensity of the primary source to the screen is "1" intensity unit, what is the intensity of the secondary source that is 3 distance units away?