Gaussian Beam focusing with lens

[faradayslaw]

Homework Statement

Find the condition needed for a gaussian beam with waist a distance d0 from a thin lens to be imaged (so its geometric image is at di from the lens as per 1/f=1/di + 1/d0) so that its new waist occurs at di, the geometric image position.

Homework Equations

q2=Aq0+B/(Cq0+d)

The Attempt at a Solution

Let zr=Rayleigh length for initial beam.
I find that the Real part of 1/q2 is Re[1/q2]=zr^2/f*(1-di/f)=1/R2. If the beam has a waist at di, then R2=Infinity, in which case Re[1/q2]=0 --> di=f, but then d0 would have to be at infinity. I am not sure if this is correct or not, so please let me know any other answers that mihgt be come across.

Thanks

 

[mark726]

for your question and for sharing your approach. I believe you are on the right track. To find the condition for the waist to occur at di, we can use the thin lens equation: 1/f = 1/di + 1/d0, where f is the focal length of the lens, di is the image distance, and d0 is the object distance. We also know that the waist of a gaussian beam is located at the Rayleigh length, zr, which is given by zr = πw0^2/λ, where w0 is the waist radius and λ is the wavelength of the beam.

Now, we can substitute the expression for zr into the thin lens equation, and we get:

1/f = 1/di + πw0^2/λd0

To have the waist at di, we need to have w0^2/λd0 = 0. Therefore, the condition for the waist to occur at di is:

d0 = ∞

This means that the object distance, d0, needs to be at infinity, which is what you had also found. This condition is necessary in order for the gaussian beam to be imaged at di with its waist at di. I hope this helps and clarifies the approach.