Incident Radiation (Part 2)
In the x y z -coordinate system, a thin solid square of side length S lies in the x y -plane with its center at the origin. An isotropic light source is positioned at ( x , y , z ) = ( 0 , 0 , 2 S ) .
What percentage of the light energy is incident on the square ( to one decimal place; i.e., 6 6 3 2 % would be entered as 66.7 ) ?
The answer is 16.7.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
If you want a similar problem, which does not give you a calculus-free backdoor, try this one .
This is a bit of a kick-self solution. In many ways it's too easy for "advanced" because once you understand the cheat it's obvious.
Log in to reply
It has a bifurcated difficulty profile, depending on how you try to solve it.
This is the most beautiful solution to any problem I've seen on this website
Are you sure the answer for the part 1 was correct?
Log in to reply
Quite confident, yes. I'll post a solution to that one.
Log in to reply
Found my mistake. I was not taking taking the dot product
I posted a solution to the other one
Could you show the calculus way I am interested
Log in to reply
Well, you can look at the solution I gave to the problem I linked. Alternatively, there are two calculus-based approaches to Part 1 of this question - one mine, and one Steven's.
Log in to reply
His is the proper one. Mine is the "engineer version". :)
Exactly my approach. Nice to know.
Please show calculus way
This looks more like geometry.
Compare the volume of the square based pyramid with base length S and an apex at the radiator to the volume of the cube with edge length S.
Imagine all you like but a light source radiates spherically. The answer to this question 1/pi.
Log in to reply
Can you elaborate?
It would help if you could give your reasons for this. We could then work out where you are going wrong...
Did this the slow calculus way and also got 1/6.
Beautiful, simple and in hindsight obvious solution, Stephen. Elmer did what I did initially (although I knew it was definitely wrong). I took the surface area of the sphere radius S/2 (=Pi x S^2) and effectively "wrapped" the square (area s^2) around it giving the ratio of the energy absorbed by the square to the whole sphere as 1/Pi or 31.8%. What I really wanted to do was workout the area of the shadow cast by the square on the sphere of radius root 2 s/2 that touches the corners of the square but knew that needed more time than I could be bothered with and knew the solution would be way more elegant - and it was. Thanks
Imagine a sphere of radius 2 S 2 . It goes around the cube whose side is shown in the question. It represents the path of the light coming off the light source.
Now imagine the aforementioned cube inside the sphere. It separates the surface of the cube into six parts, making the light absorbed by one face 6 1 0 0 % of the total.
Just cover the light with cube (light at centre)
Now remove 5 sides
All sides gets equal intensity so answer is clearly 💯/6
This could be done with calculus, but here's an easier way.
1) Imagine a hollow cube with an isotropic radiator at the center
2) It is obvious that each side of the cube absorbs 6 1 of the total power
3) Now take away all but one side, and we're left with the problem as stated. So the answer is 6 1 ≈ 1 6 . 7 %