Irradiance of sun in an image?

[carnivalcougar]

Homework Statement

What will be the irradiance (power per unit area) in the image, if the Sun's irradiance at the sea level is I = 500W/m^2

Lens is 200mm f
Sun is 150x10^9 m away
Diameter of sun is 1.4x10^9 m

Homework Equations

The Attempt at a Solution

I know that at the lens the irradiance will be the same i.e. 500W/m^2 but at the image the irradiance will be magnified. I'm not sure how to get there though.
I know an image of the sun will be .2m from the lens and the diameter of the image will be .00187m because M=di/do = .2m/150x10^9 = 1.333X10^-12 and 1.333x10^-12 multiplied by the diameter of the sun is .000187m.

 

[haruspex]

What's the total power falling on the lens? What's the total power in the image?

 

[carnivalcougar]

I'm not sure how to find the diameter of the lens which is what I need to find the total power falling on the lens

 

[carnivalcougar]

I think that the radius of the lens is equal to 2f which would make the radius .4m. Therefore, the area of the lens is .50265m^2. Multiply this by 500 and this tells you the power on the lens which is 251.327W. The area of the image is 1.099X10^-7m^2 which is from ((.000187m)^2(pi)). Therefore, 251.327W also falls on this same area and dividing 251.327W by the area of the image (1.099X10^-7m^2) will give you the power per m^2 which is 2.29X10^9 W/m^2

Is this correct?

 

[haruspex]

I'm not sure how to find the diameter of the lens which is what I need to find the total power falling on the lens

Good point. Then I don't see how you can answer the question. carnivalcougar gives a way to find an upper bound, since the focal length sets a limit on the diameter of the lens. But it will depend on the refractive index.