Incident Radiation (Part 2)

In the x y z xyz -coordinate system, a thin solid square of side length S S lies in the x y xy -plane with its center at the origin. An isotropic light source is positioned at ( x , y , z ) = ( 0 , 0 , S 2 ) (x,y,z) = \left(0,0,\frac{S}{2}\right) .

What percentage of the light energy is incident on the square ( ( to one decimal place; i.e., 66 2 3 % 66\frac{2}{3}\% would be entered as 66.7 ) ? )?


The answer is 16.7.

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3 solutions

Steven Chase
Jul 2, 2017

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 1 6 \frac{1}{6} 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 1 6 16.7 % \frac{1}{6} \approx 16.7 \%

If you want a similar problem, which does not give you a calculus-free backdoor, try this one .

Mark Hennings - 3 years, 11 months ago

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.

Malcolm Rich - 3 years, 11 months ago

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It has a bifurcated difficulty profile, depending on how you try to solve it.

Steven Chase - 3 years, 11 months ago

This is the most beautiful solution to any problem I've seen on this website

Rishy Fishy - 3 years, 11 months ago

Are you sure the answer for the part 1 was correct?

Shivansh Kaul - 3 years, 11 months ago

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Quite confident, yes. I'll post a solution to that one.

Steven Chase - 3 years, 11 months ago

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Found my mistake. I was not taking taking the dot product

Shivansh Kaul - 3 years, 11 months ago

I posted a solution to the other one

Steven Chase - 3 years, 11 months ago

Could you show the calculus way I am interested

Mazen Elkhouly - 3 years, 11 months ago

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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.

Mark Hennings - 3 years, 11 months ago

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His is the proper one. Mine is the "engineer version". :)

Steven Chase - 3 years, 11 months ago

Exactly my approach. Nice to know.

Aarabdh Tiwari - 3 years, 11 months ago

Please show calculus way

ARJUN KUMAR - 3 years, 10 months ago

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See Part 1

Steven Chase - 3 years, 10 months ago

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.

Akeel Howell - 3 years, 2 months ago

Imagine all you like but a light source radiates spherically. The answer to this question 1/pi.

elmer pryor - 3 years, 11 months ago

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Can you elaborate?

Steven Chase - 3 years, 11 months ago

It would help if you could give your reasons for this. We could then work out where you are going wrong...

Mark Hennings - 3 years, 11 months ago

Did this the slow calculus way and also got 1/6.

Matt Williams - 3 years, 11 months ago

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Same here, just as a double check

Steven Chase - 3 years, 11 months ago

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

Chris Kowalski - 3 years, 10 months ago
Brian Lamptey
Jul 13, 2017

Imagine a sphere of radius S 2 2 \dfrac{S\sqrt{2}}{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 100 6 % \dfrac{100}{6} \% of the total.

Yash Ghaghada
Mar 2, 2018

Just cover the light with cube (light at centre)

Now remove 5 sides

All sides gets equal intensity so answer is clearly 💯/6

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