A. What is the peak wavelength for AM0? What temperature T corresponds to this peak wavelength for a blackbody source? Assuming this T as an estimate of the sun’s surface temperature, calculate: i) surface irradiance, ii) irradiance at one astronomical unit (distance of sun-to-earth)

A) T = 5510 K , B)   I = 5,226 10⁷ W / m² , C)   I₂ = 1128 W / m²

Explanation:

A) The mass of air is defined as the ratio between the shortest path that sunlight must pass to reach the planet's surface and the length of the beam, is called AM

      Am = 1 / cos θ

The AM0 value corresponds to solar radiation in the outer part of the Earth's atmosphere.

The peak of this emission is the peak that emitted from the sun

       λ = 526 nm

To find the temperature that corresponds to this emission we use the Wien displacement law

       λ T = 2,898 10⁻³

      T = 2,898 10⁻³ / 526 10⁻⁹

      T = 5510 K

i) The radiance on the surface of the sun is

           I = P / A

We can calculate the potency by Stefan's law, for a black body

         P = σ A e T⁴

         P / A = σ e T⁴

The σ constant is value 5,670 10⁻⁸ W / m²K⁴, we will assume that the Sun emits as a black body, so e = 1

            I = sig T⁴

            I = 5,670 10⁻⁸  5510⁴

            I = 5,226 10⁷ W / m²

ii) the irradiation at a distance of 1 ua (1,496 1011 m)

Let's use the relationship

           P = I A

           I₁ A₁ = I₂ A₂

           I₂ = I₁ A₁ / A₂

The area of a sphere is

          A = 4π R²

Let's replace

         I₂ = I₁ (r₁ / r₂)²

  Index 1 corresponds to the sun and the index to Earth that is an astronomical unit

        r₁ = 6.96 10⁸m        (Sun radius)

        r₂ = 1,498 1011 m    (Earth-Sun distance)

Calculous

       I₂ = 5,226 10⁷ (6.96 10⁸ / 1,498 10¹¹)²

       I₂ = 1.1281 10³ W / m²

       I₂ = 1128 W / m²

snell's law

refractve index = sin incidence / sin refraction