 #1
baseballfan_ny
 91
 23
 Homework Statement:

How many commercially available light bulbs of 100W each placed at 1m away from the
atmosphere’s edge would generate an irradiance of 1370 W/m^2 on an area of 1m2?
 Relevant Equations:
 Solid angle = spherical volume integral/rho^2
The actual problem can be found as #2 on this link: https://ocw.mit.edu/courses/mechani...sspring2009/assignments/MIT2_71S09_ups1.pdf
I rewrote the problem above with the solar irradiance data that they give.
My interpretation is of a square 1 m x 1 m plane sitting behind an imaginary sphere. So the angle ##\alpha = \arctan(0.5)##. The angle swept by phi in the vertical direction should go from ##\alpha## to ##+\alpha## and the angle swept by theta radially in the xy plane should also be the same given the square shape.
The solid angle of a sphere is ##4\pi##, so the ratio of the solid angles I need is $$ \frac {\Omega} {4\pi} = \frac {1370} {100 N} $$
where N is the number of lightbulbs.
Now for calculating the solid angle ##\Omega## I am getting tripped up.
$$ d\Omega = \sin(\phi)d\phi d\theta $$
$$ \Omega = \int_{\theta = \alpha}^{\alpha} \int_{\phi = \alpha}^{\alpha} \sin(\phi) \, d\phi \, d\theta $$
$$ \Omega = 2\alpha * (\cos(\alpha)   \cos(\alpha)) $$
$$ \Omega = 2\alpha * (\cos(\alpha) + \cos(\alpha)) $$
$$ \Omega = 0 ?!? $$
I know that can't be right
Ω
I rewrote the problem above with the solar irradiance data that they give.
My interpretation is of a square 1 m x 1 m plane sitting behind an imaginary sphere. So the angle ##\alpha = \arctan(0.5)##. The angle swept by phi in the vertical direction should go from ##\alpha## to ##+\alpha## and the angle swept by theta radially in the xy plane should also be the same given the square shape.
The solid angle of a sphere is ##4\pi##, so the ratio of the solid angles I need is $$ \frac {\Omega} {4\pi} = \frac {1370} {100 N} $$
where N is the number of lightbulbs.
Now for calculating the solid angle ##\Omega## I am getting tripped up.
$$ d\Omega = \sin(\phi)d\phi d\theta $$
$$ \Omega = \int_{\theta = \alpha}^{\alpha} \int_{\phi = \alpha}^{\alpha} \sin(\phi) \, d\phi \, d\theta $$
$$ \Omega = 2\alpha * (\cos(\alpha)   \cos(\alpha)) $$
$$ \Omega = 2\alpha * (\cos(\alpha) + \cos(\alpha)) $$
$$ \Omega = 0 ?!? $$
I know that can't be right
Ω