00:01
For this question, we're going to need to find the power per unit area at a specific wavelength.
00:07
So let's start with power as a function of our radiance.
00:12
So we know that if we want power at some specific lambda interval, so some delta lambda, this is going to be equal to our radiance at some lambda and t times the interval that we're working with, the delta lambda, times the area.
00:39
So now we can go ahead and rearrange this equation to get this in terms of power per unit area.
00:48
So we're going to have p of our delta lambda per unit area is equal to our radiance times our delta lambda.
00:59
Now let's go ahead and substitute in our equation for radiance, which we know to be 2 pi hc squared over top of lambda, so the wavelength, to the power of 5 times e to the exponent of hc over lambda kt, and then that e term is minus 1, and that is according to the planck blackbody radiation curve that was formulated in the year 1900.
01:54
And then we can multiply that by the shift in our wavelength, delta lambda, and that gives us power per unit area...