- #1

baseballfan_ny

- 92

- 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...s-spring-2009/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