Answer:
A)Ω = 7.8 × 10^−5 steradians.
B) TE = 5800K
C) fλ(λ1) = (π
^2
) /ΩBλ(T)
Explanation:
A) First of all, if we assume that the Sun emits isotropically at a luminosity (L⊙) , the flux at a given
distance R from the sun would be f(d) = L⊙/ (4πd^2)
The ratio of flux at the solar photosphere to the flux at the Earth’s atmosphere would be: F⊙/{f(d⊙)} = (R⊙)^2 / (d⊙)^2
Now if we think of this relationship of the flux and the earth as a conical pattern, we'll deduce that the solid angle subtended by the sun at Earth’s surface to be;
Ω = π[(R⊙)^2 / (d⊙)^2]
Combining this with the ratio earlier gotten, well arrive at;
F⊙ = {f(d⊙ )π} /Ω
Now let's express The radius of the sun (R) in terms of its angular diameter (2α) and this gives;
R⊙ ≈ αd⊙
Now combining this with the equation for Ω earlier, we get;
Ω ≈ πα^2
So, = π((0.57/2π) /180)^2 = 7.8 × 10^−5
steradians.
B) from Stefan-Boltzmann Law,
F⊙ = σ(TE)^4
From the beginning, we know that;
F⊙ = {f(d⊙ )π} /Ω
And so replacing that in the stephan boltzmann law, we get ;
{f(d⊙ )π} /Ωσ = (TE)^4
So, (TE)^4 = {π (1.4 kWm^(−2))} / [(7.8 × 10^(−5
) steradians x (5.66961 × 10^(−8))]
In stephan boltzmann law, σ = 5.66961 × 10^(−8)
And so, TE is approximately 5800K.
C) In order to relate fλ(λ1) with T, let's assume the sun’s surface to be an isotropically emitting blackbody, i.e its specific intensity is Iλ = Bλ(T). Hence, the flux at Sun’s surface for a given wavelength would be;
Fλ(λ1) = πBλ(T)
Now, if we combine this with the expression of F⊙ gotten earlier, well get the relation;
fλ(λ1) = (π
^2
) /ΩBλ(T)
" .