I was starting to study optical fibers and I came to the chapter: Classifications of optical fibers. Here I read about singlemode and multimode fibers, regarding the latter he explained the difference between step-index and graded-index. My question is: why do the signals sent in the step take different times while in the graded they don't? I did not understand very well the difference between these two refraction profiles.
Optics – Differences Between Step-Index and Graded Index Fiber
fiber opticsopticsrefraction
Related Solutions
As you've observed in your edit, if you flip the orientation of the beamsplitter such that the thick part is now on the back side, then you still need to use the compensator plate in the other arm.
You might wonder why a compensator plate is used at all. In an interferometer which uses monochromatic light (i.e. a laser) you could just shift the position of one of the end mirrors to compensate for the extra optical path length picked up in the thick part of the beamsplitter. Real laser beams though have a finite spatial extent which changes as they propagate due to diffraction. When this beam passes through a material with a non-unitary index of refraction the mode changes more slowly. This effect is opposite to the change in phase while propagating through the medium. I.E. $$ d_{OPL}=n\ d \qquad\text{but}\qquad d_{BS}=\frac{d}{n} $$ where $d$ is the thickness of the glass plate, $d_{OPL}$ is the equivalent in-vacuo distance for the phase change, and $d_{BS}$ is the equivalent in-vacuo distance for the beam size change. Most practical interferometers use well-collimated beams whose size doesn't change much over the size of the interferometer so this effect is often unimportant.
If you are using a more broadband source in your interferometer, such as a lightbulb or a discharge lamp, then there is another advantage of using the glass plate. As pointed out in @User58220's comment, the dispersion in the material means that the different frequencies in your source will each see a slightly different length in the substrate of the beamsplitter. Using the same material in the other arm allows you to add exactly the same amount of optical distance for all of the different wavelengths in your source.
Use Fermat's prinicple. Let the thickness of the lens be denoted by $d$, then the optical path length from axial point $\mathcal P$ to another axial point $\mathcal Q$ on the other side of the lens is $$OPL[{\mathcal P \mathcal Q}]=\sqrt{p^2+r^2}+n(r)d+\sqrt{q^2+r^2}\\ \approx p+\frac{r^2}{2p} + n(r)d + q+\frac{r^2}{2q}$$ For imaging this must be independent of $r$, that is, $$ \frac{r^2}{2}\big(\frac{1}{q}+\frac{1}{p}\big) + n(r)d =\frac{r^2}{2f}+n(r)d=n_0$$ where $\frac{1}{q}-\frac{1}{p}= \frac{1}{f}$ with $$n(r)=n_0-\frac{r^2}{2f}$$
Best Answer
In a ray optics model, if you launch light into a step index multimode fiber at a high angle to the axis, it will bounce back and forth between the "walls" of the core rather than travel straight along the axis. So it will take longer to reach the other end of the fiber than a ray launched along the axis.
If you launch a ray at an angle into a graded-index fiber, it will also bounce back and forth between the "walls". But the higher the angle, the more time it will spend near the walls, where the index is lower and it can propagate faster. If you launch a ray along the axis, it will mostly propagate in the high-index material at the axis, and thus propagate slower. If you get the index profile just right (a perfect parabolic profile) the two effects will balance and the ray launched on axis will take just as long to reach the other end as the one launched at an angle.