to find viewing angle of an LED

[nikhila]

Sir

2 identical light soure are kept at a distance 7.5mm (dimension of the source is 3*2)

I need a uniform illumination (meeting of light from the sources at a distance above the source) at 30 mm above.
Then what should be the angle through light is passing out of the source.

hope u will hepl me
regards
naikhila

 

[Euler]

If what you are asking for translates into the following picture:

Then simply take the inverse tangent (tan^-1) of the given sides, and you will get the angle, which I have marked 'Ø'.

What did you mean by "the dimension"?

 

[Gene]

If I'm reading the question correctly the centers of the (round?) LEDs are sqrt(9/pi)+7.5 = 10.9 mm apart.
The angle between the midpoint of the line between the LEDs and the center of the LEDs is
arctan((10.9/2)/30) = 10.3 degrees. That should be the required angle.
If you mean they are 3 by 3 mm squares the distance becomes 10 mm and the angle becomes arctan(5/30) = 9.5 degrees.
The angles are from the vertical.

 

[nikhila]

thanks

Then viewing angle should be 2 theta.
I done it in the same way and got 2 thata as 14.25.but i was not confident.

By dimension i ment to say the length and breadth of the source.
thanks alot

 

[Gene]

Since I'm outvoted two to one, I'll just say that the only reason I can think of for including the dementions of the LED is so you can find the center-to-center distance.
BTW: Did you notice that you changed "angle light is passing out of the source" to "viewing angle"? Better make sure before you publish.
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Gene