Plane Mirror Problem: Solve Intensity Using I = Ps/(4*PI*r^2)

[DANIELLYMA]

I'm having problems doing the problem below. The equation in the hint is: I = P_s/(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.

 

[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)

 

[DANIELLYMA]

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

 

[tiny-tim]

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

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

 

[DANIELLYMA]

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

 

[tiny-tim]

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

I don't follow … r is distance.

 

[Redbelly98]

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