Timeline for Transmission of Gaussian Beam Through Graded-Index Slab
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2012 at 16:13 | vote | accept | John Roberts | ||
| Oct 2, 2012 at 15:54 | comment | added | Ondřej Černotík | let us continue this discussion in chat | |
| Oct 2, 2012 at 15:48 | comment | added | John Roberts | I am new to this concept. Could you briefly show me how this would be done? | |
| Oct 2, 2012 at 15:36 | comment | added | Ondřej Černotík | To get $1/R_2$ in terms of those parameters, you just take the real part of $1/q_2$. | |
| Oct 2, 2012 at 15:31 | comment | added | John Roberts | I know that $\frac{-2i}{k W_2^{2}}$ can be expressed in terms of $\alpha, d, W_0, \lambda_0$. However, to get the full width function, I need to express the real part ($\frac{1}{R_2}$) in those terms as well, or get rid of it entirely. Could you please help me with that? | |
| Oct 2, 2012 at 14:08 | comment | added | Ondřej Černotík | Because all the parameters (except $q$) are real, you can split $1/q_2$ to real and imaginary part. The real part is then equal to $1/R_2$ while the imaginary part is $-2/(kW_2^2)$, which will then be expressed only in terms of $\alpha, d, W_0, \lambda_0$. | |
| Oct 2, 2012 at 13:41 | comment | added | John Roberts | The beam width function that I get must be a function of d, which means I cannot include $\frac{1}{R_2}$ in it. I then have to express $\frac{1}{R_2}$ in terms of $\alpha$, d, $W_0$, $\lambda_0$, or get rid of entirely. This is the part I'm unsure about. Is it possible that the beam leaves at its waist as well, allowing us to eliminate $\frac{1}{R_2}$? | |
| Oct 2, 2012 at 11:41 | history | edited | Ondřej Černotík | CC BY-SA 3.0 |
added 2 characters in body
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| Oct 2, 2012 at 11:41 | comment | added | Ondřej Černotík | $i$ and $-1/i$ are the same, so both expressions are correct, expect fot the square of $W_0$ which I missed. Now, you can get $q_2$ from $q_1$ using the ABCD transformation of complex beam radius as above and then using the definition $1/q_2 = 1/R_2 - 2i/(kW_2^2)$, you can determine both the radius of curvature and the beam width lleaving the slab. | |
| Oct 1, 2012 at 23:57 | comment | added | John Roberts | Thank you for your reply. Wouldn't it be $q1=-kW_0^{2}/(2i)$ though? And this would just give us an expression for q1, from which we can get an expression for q2. However, we still need to substitute out for q2, and to do this, we need to get the radius of curvature for the beam leaving the slab. I'm still not sure how to do this. | |
| Oct 1, 2012 at 20:09 | history | answered | Ondřej Černotík | CC BY-SA 3.0 |