Formula For Calculating Field Angle (Beam Angle) of Light?

[meeshu]

There are actually several questions.

The formula for calculating the "beam angle" of light (emitted from lights bulbs, flashlights etc) is -

α = 2 arcCos ( 1 - Lm ÷ ( 2 π Cd ))

Where, α = beam angle in degrees
Lm = luminous flux (Lumens)
Cd = luminous intensity (Candela)

The above formula is simply a rearrangement of the equation -

Lm = 2 π Cd ( 1 - Cos (α ÷2) )

The beam angle formula above indicates the angle of light beam where the beam intensity falls off to 50% of the beam maximum intensity (apparently).

First question:
How is the equation Lm = 2 π Cd ( 1 - Cos (α ÷2) ) derived?

Second question:
How do we know that this equation applies only to 50% beam intensity?

There is another angle of light known as the "field angle", where the light beam angle is determined where the beam intensity falls to 10% of the maximum intensity. The field angle is greater than the beam angle.

Third question:
What is the formula for calculating light field angle?

Thanks for any and all constructive comments!

 

[mcastillo356]

Hi, meeshu, I'm afraid I cannot help you, haven't got idea. The Internet doesn't help you? YouTube, for example. Have you got PC, or just a mobile? Have you got a library near? Sorry, I cannot provide anything but questions. How did you acquired the background? Wouldn't you prefer to tackle one question, just to start moving?
Greetings!

 

[meeshu]

Thanks for your comments.

I had already searched the internet before posting this topic. I didn't get satisfactory results from the searches at that time.

After more searches I believe the first question has been answered, courtesy of information from this
link here !

But from another equation,

Lumen = Candela x Steradian, where the solid angle steradian equals 2 π ( 1 - Cos (α ÷ 2) )

So the lumen equation becomes -

Lm = 2 π Cd ( 1 - Cos (α ÷2) ) as per the same equation mentioned in my first post.

But my second question as to why this equation only relates to beam intensity falling off to 50% instead of the entire beam intensity falling off to 0, has not been answered!

It is not clear why the equation only applies to 50% beam intensity?? That seems a bit odd!

In summary, to date, there are two remaining questions yet to be answered -

Why does the equation only apply up to 50% beam intensity?

And, what is the equation for calculating light beam "field angle" where the beam intensity falls down to 10% of the initial (maximum) beam intensity?

Maybe these questions should be in a "photometric" or similar sub-forum, but I don't think there is such a sub-forum here(?)