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a) Overlap integral given by \[ \eta=\int_{-\infty}^{\infty} \Psi_{m^{'}} ^{output} \Psi_{m}^{input} dx\] \\  b) Using the mode solver for this part. I wanted to see how modes would look after $LP_{01}$, not single moded. So I went back to the paper and looked at the equation for the diameters that allow for single mode operation, $D< \frac{2.4 \lambda}{\pi \sqrt{n_0^2-n_1^2}}$, where $n_1$ is 1 for air, and $n_0$ is the index of refraction for the medium in use. From the tong paper, the second page of the paper, or pg.817 of the nature journal it was published in, index of refraction for Silica is said to be $n_0=1.46$. This gives $NA=1.063$, using $\lambda=633 nm$, the max diameter is then given to be 454.6 nm for single mode operation. So I will put diameters larger then this for non single modes. Using 600 nm for the diameter, Ive included diameter.\\  c) I believe this is due to when the diameter of the wire is decreased below the wavelength its supposed to be guiding, more of the light is guiding outside the wire as a surface wave. So for 1550 nm , looking at the graph for loss, starting at wire diameters of 1200 nm we are already operating below  the images wavelength we want to guide so as the diameter decreases , more light is outside the wire leading to more loss. Compared to the 633 nm wavelength, you can see the increase  in figure 1 and figure 2. loss occurs when operating below 633 nm diameter wire but we are lower loss from 1200 nm until we get to 633 nm\\  d) To go along with this, the loss mechanism is from surface contamination. The silica wires are't perfectly uniform. By virtue of that, when light is guided by surface waves it is more susceptible to the surface contaminations.