Ne W2
1 0 KI. Calculate the luminous inten
Uislunce of the
luminous intensity and the
[HSEB 20651
U
total luminous flux from the sun
Solution:
Here, illuminance (I) = 1.5 x 10° lumen/m²,
Distance (T) = 1.5 x 10 km =1.5 x 108 m
Let L be the luminous intensity of the sun, then
or, L= TXT =1.5 x 10°x (1.5 x 108)2
= 3.375 x 102 cd.
Total luminous flux from the sun = 410 I
= 41 x 3.375 x 1021 = 4.24 x 1022 /m
Example 3. If a source has luminous intensity of 400 cd, (a) how much
luminous flux is emitted by the source? (b) what is the illuminance at a point
on the table which is 1.6 m directly below the source?
Solution:
Here, Luminous intensity (L) = 400 cd

3.82746e+26 watts

Explanation:

There are two ways to solve this problem. One way is to use the equation

L = 4πσR²T⁴

where

L = the sun's bolometric (all-spectrum) luminous power

σ = 5.670374419e-8 W m⁻² K⁻⁴ = the Stefan-Boltzmann constant

R = 6.957e+8 meters = the sun's radius

T = 5771.8 K = the sun's effective temperature

You find that

L = 3.82746e+26 watts

The other way to solve the problem is to use the Planck integral for radiant flux.

L = 4π²R ∫(v₁,v₂) 2hv³/{c² exp[hv/(kT)]−1} dv

where

h = 6.62607015e-34 J sec

c = 299792458 m sec⁻¹

k = 1.380649e-23 J K⁻¹

v₁ = 0 = frequency band lower bound, in Hz

v₂ = ∞ = frequency band upper bound, in Hz

You find, once again, that

L = 3.82746e+26 watts

The advantage of using the Planck integral becomes clear when you want to calculate the sun's luminous power only in a specific band, rather than across the entire spectrum. For example, if we do the calculation again, except that we use

v₁ = 4.1e+14 = frequency band lower bound, in Hz

v₂ = 7.7e+14 Hz = frequency band upper bound, in Hz

restricting ourselves to the visible spectrum. We find that

L (visible) = 1.56799e+26 watts

So the fraction of the sun's luminosity that is in the visible spectrum is

L (visible) / L = 0.4096686