Irradiance of sun in an image?

In summary, the irradiance in the image will be magnified from the Sun's irradiance of 500W/m^2 at sea level. To find the total power falling on the lens, we need to know the diameter of the lens, which can be estimated using the focal length as an upper bound. However, the exact diameter will depend on the refractive index. The total power in the image can be calculated by multiplying the lens area by 500W/m^2 and dividing by the image area.
  • #1
carnivalcougar
40
0

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.
 
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  • #2
What's the total power falling on the lens? What's the total power in the image?
 
  • #3
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
 
  • #4
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?
 
  • #5
carnivalcougar said:
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.
 

1. What is irradiance of the sun?

Irradiance of the sun refers to the amount of solar energy received per unit area on a surface. It is a measure of the intensity of sunlight and is typically expressed in watts per square meter (W/m²).

2. How is irradiance of the sun measured?

Irradiance of the sun is typically measured using a radiometer, which is a device that detects and measures electromagnetic radiation. The radiometer is pointed towards the sun and the intensity of the sunlight hitting the sensor is recorded.

3. What factors affect the irradiance of the sun in an image?

The irradiance of the sun in an image can be affected by a variety of factors including the time of day, weather conditions, and atmospheric conditions. The position of the sun in the sky and the presence of clouds or other obstructions can also impact the irradiance.

4. Why is measuring irradiance of the sun important?

Measuring irradiance of the sun is important for a variety of reasons. It helps us understand the amount of solar energy available in a given area, which is important for renewable energy applications. It also provides valuable data for climate and weather studies, as well as for monitoring changes in the Earth's atmosphere.

5. How does the irradiance of the sun vary across the globe?

The irradiance of the sun varies across the globe due to a number of factors such as latitude, altitude, and weather patterns. Generally, areas closer to the equator receive more sunlight and have higher irradiance levels, while areas closer to the poles receive less sunlight and have lower irradiance levels.

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