# Spherical mirror area according to the amount of light

1. Sep 2, 2013

### ex3mist

Hello,
I need some information about spherical mirrors that I can't find in internet or this forum.
1. How to calculate the amount of light that is focused in the mirror's focus point depending on the mirror's area and the amount of light emited by the source?
2. If that light is reflected by another plain mirror to another point, is there any loss of light? I mean loss of efficiency.
The purpose of my questions is if I want certain amount of light, how can I calculate the spherical mirror's area and to take into account any losses?

Thanks!

2. Sep 2, 2013

### Staff: Mentor

Other than the surface area of the mirror all I think you need is how well the mirror reflects the light. There are no perfect mirrors and adding another to your optical system will always reduce the amount of light by at small amount. Or more if you use low quality mirrors.

3. Sep 2, 2013

### ex3mist

Well, of course the quality of the mirror is very important, but I consider it would be used high quality mirrors. What bothers me most is is there an equation that defines the dependence between the area of the mirror and the amount of light in its focus point? If I want certain amount of light, how can I determine how big the mirror has to be?

4. Sep 2, 2013

### Staff: Mentor

How much energy per unit time do you want to deliver to an object at the focal point?

How much energy does your light source deliver to a unit area of the mirror per unit time?

Think about the physical significance of these two quantities for a moment, and it will be clear ho wto calculate the required area of the mirror from them.

5. Sep 2, 2013

### Low-Q

@ ex3mist:
What do you need a spherical mirror for? Wouldn't it be better with a parabolic mirror instead? Parabolic mirrors do focus parallell incoming light into a single point. Spherical mirrors don't.

I ask because you mention focus point. A spherical mirror does not have a defined focus point, a parabolic mirror does.

Vidar

6. Sep 2, 2013

### ex3mist

@Nugatory:
I just want an equation. If there is one... The source is the Sun. The light varies depending on the weather.

@Low-Q:
I'm not sure about the terms, maybe you're right. I just want to use Sun light and to focus it.

7. Sep 10, 2013

### Low-Q

Isn't it approx 6 kW per square meter at equtor? Mayby I don't remember correctly.
Focus that energy into a square centimeter. You might get hurt in your fingertip if you put it there :-) The energy will still be approx 6 kW there - minus loss from the none-perfect mirror. The size of the spot depends on the size and distance of the sun, and what the focal length are. The focal length is the only possible variable. Sorry for not having an equation for you.

Vidar.

8. Sep 10, 2013

### ex3mist

Thanks Vidar, I didn't know about the 6 kW per square meter. That gives me some ideas. Maybe that information is better. ;) Can I read more about it somewhere? Can you give a link, please?

9. Sep 10, 2013

### cjl

That number is a little high - if I remember right, it's about 1kW per square meter below the atmosphere, and about 1.2kW per square meter in space (latitude doesn't matter, as long as the collector is directly facing the sun)

10. Sep 11, 2013

### Low-Q

Quote from Wikipedia (I overrated that energy by some factor):
"The total amount of energy received at ground level from the sun at the zenith is 1004 watts per square meter, which is composed of 527 watts of infrared radiation, 445 watts of visible light, and 32 watts of ultraviolet radiation."

So there you go. Here is the link: http://en.wikipedia.org/wiki/Sunlight

Vidar