Laser Beam Divergence: Calculating Total Output Power and Spot Diameter

[Ikaros]

Homework Statement

A 50W lamp radiates isotropically and has a spectral width of 750 nm and a centre frequency of 600 nm.

How much of its total output power is emitted into a solid angle of 10^-6 Sr?

What is the diameter of the spot that a beam subtending this solid angle makes on a perpendicular screen placed 1 m from the source?

Homework Equations

Full sphere 4*Pi Sr

The Attempt at a Solution

For the first part of the question, I took the approach of taking the ratio of the solid angle to a full sphere and applying it to the total power radiated from the lamp (10^6/4*pi)*50W, which gave me ~4microwatts. Have I taken the right approach?

For the second part of the question, I thought I'd first have to calculate the beam divergence which would lead me to the diameter of the spot, but I'm not given the diameter of the source. I'm assuming the spectral band comes into play here, but I'm hoping someone can give me a nudge in the right direction.

 

[ehild]

You answered the first question correctly.
As for the second question you can take the light source point-like.
Why do you call it laser? It is said radiating isotropically and having a spectral band of 750 nm around 600 nm, so it radiates in the whole visible range. The divergence of a laser beam is very narrow and also the bandwidth of its radiation. This is just an ordinary lamp. ehild

 

[Ikaros]

Oops, that is an error in the title of the thread and I'll blame it on the fact the question is from a tutorial on laser physics and that the first question in full is actually a comparison between the power delivered in a solid angle of a lamp and a laser - which is now more clear.

For the second question, I intend to use \theta = 2\arctan \left( {\frac{D}{{2L}}} \right)
in an effort to solve for the diameter of beam spot. I hope you let me know if I'm going down the wrong path.

 

[ehild]

Take care, the solid angle is not the same as \vartheta. It is related to the solid angle Ω as

Ω=2\pi(1-cos\vartheta)

(http://en.wikipedia.org/wiki/Solid_angle)

This relation can be written also in the form

Ω=4\pi sin²(\vartheta/2).

Subbing Ω=10^-6, you get \vartheta. Take care, \vartheta is in radians. As the solid angle is very small, sin(\vartheta/2)=\vartheta/2.
The relation between D, L and \vartheta is correct.

tan(\vartheta)=D/(2L).

ehild

 

[Ikaros]

That's great. Thanks again for the support.