# How do you calculate heat flux (Kw/m2) at the focal point of a mirror? [duplicate]

can anyone help me to determine the heat flux (Kw/m2) on a focal point of a parabolic dish having a diameter of 1.5 meter and a focal length 60 cm ??? please awaiting your soonest reply for my senior project :( Regards

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## marked as duplicate by Waffle's Crazy Peanut, John Rennie, Qmechanic♦Apr 9 '13 at 22:22

Possible duplicate: physics.stackexchange.com/q/59438/11062 – Waffle's Crazy Peanut Apr 9 '13 at 15:31
Also, a homework-kinda question..! – Waffle's Crazy Peanut Apr 9 '13 at 16:08

In any practical application, the flux measured is the average over the area of the measuring apparatus.

In the limit of a perfect dish, an infinitely far away source, and no diffraction, then the flux would be infinite at a single point and zero everywhere else in that plane. But all of these conditions are violated, and so the flux is spread out over a region in the focal plane.

The best thing to do is measure it experimentally. If you have $1~\mathrm{kW}$ of power hitting a receiver that is $2~\mathrm{cm} \times 2~\mathrm{cm}$, then you know the average flux falling on the receiver is $2500~\mathrm{kW}/\mathrm{m}^2$.

One could work out the diffraction-limited value, but making a dish that large be that perfect for visible light is very hard, so I doubt your flux will reach this theoretical maximum.

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Site rules rule out a direct calculation I'm afraid but here are some clues as to how I would go about it:

Use the underlying principles of optics and energy conservation to equate the initial energy to the final energy...

Remembering we are dealing with an energy density here and that no finite optics can focus to an infinitesimal point...

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