# Plane Mirror Problem

1. Mar 18, 2009

### DANIELLYMA

I'm having problems doing the problem below. The equation in the hint is: I = Ps/(4*PI*r^2) which I have not clue how it relates to this question except for that it's an equation for intensity.

I know that there will be a image produced a distance d behind the mirror, so there will be another "light source" 3d away from the screen.

Last edited: Mar 18, 2009
2. Mar 18, 2009

### Ofey

We define intensity as

$$I=\frac{P_{power}}{A_{rea}}$$

If we assume that the source is pointlike, the intensity will spread like a sphere from the source. $$4\pi \ r^2$$ is the area of a sphere with the radius r. The intensity at the distance r from a pointlike source is thus

$$I=\frac{P}{4\pi \ r^2}$$

Does that help?

(I noticed that in your formula you don't have the radius squared. Are you sure that it isn't squared in your book? This is how I would interpert the formula, might be I am wrong)

3. Mar 18, 2009

### DANIELLYMA

oops thanks for pointing out the typo but I still don't understand what power has to do with this problem.

4. Mar 18, 2009

### tiny-tim

Hi DANIELLYMA!

It doesn't really matter what power has to do with the problem …

you're only asked to say how the intensity changes

use the formula to calculate by what proportion it goes up because of the "extra" source

5. Mar 18, 2009

### DANIELLYMA

r is the radius of the source, how do I take into account the the 3d distance from the screen?

6. Mar 18, 2009

### tiny-tim

I don't follow … r is distance.

7. Mar 18, 2009

### Redbelly98

Staff Emeritus
Yes, r is "the 3d distance from the screen".