Calculating Resonator Stability with Lens & Mirror

[CptXray]

Homework Statement

I have a laser cavity with one flat mirror ##R_{1} = \infty,## one positive lens with the focal length ##f_{lens}=50 \, mm## and curved mirror on the end of curvature ##R_{2}=100 \, mm.## The total length of the resonator is fixed and is equal to ##L = 1000 \, mm.## For what values of ##d## is the resonator stable? What is the distance d for which the waist size between the lens and mirror is the smallest? What is the size of the beam waist in this case? Assume ##TEM_{00}## mode.

Relevant Equations

Focal length of curved mirror:
$$f_{curved} = \frac{R_{2}}{2}.$$

I guess the resonator will be stable when both focal lengths of lens and curved mirror meet. The curved mirror is fixed and it's focal length:
$$f_{curved} = \frac{R_{2}}{2} = 50 \, mm.$$ Then the d should be: $$d=f_{lens}+f_{curved} = 100 \, mm.$$ I think that's also the distance for which the waist size is the smallest, but I don't really know how to get it's exact size. If someone who knows what's going on here could give me a hint or textbook where this is explained I'd be really grateful.

 

[NateD]

A:If the curved mirror is a parabolic mirror, then its focal length is equal to its radius of curvature. So, you have correctly calculated the distance between the lens and the mirror as $d=100$ mm. The beam size at the waist is given by the formula$$w_0=\frac{\lambda}{\pi}\sqrt{\frac{2f}{1+f/f_r}}$$where $\lambda$ is the wavelength of the light, $f$ is the focal length of the lens and $f_r$ is the radius of curvature of the curved mirror.Using your values, you get$$ w_0 = \frac{\lambda}{\pi}\sqrt{\frac{200}{1+200/50}} \approx 0.38 \,\lambda $$