lensesopticsrefraction
According to the Lensmaker's formula,
$$\frac{1}{f} = (μ-1)\left(\frac{1}{R_1} – \frac{1}{R_2}\right)$$
Apparently, focal length $f$ is inversely proportional to the refractive index of the medium $μ$. Since, Optical Power $P = \frac{1}{f}$. This should imply that optical power is directly proportional to the refractive index. However, today I got told by my teacher that these two are independent. Where have I gone wrong in my reasoning? Is the fact provided to me correct and if yes, how?
Using $f$ for frequency, we have $λf=c$, which gives us $n={c\over v}={\lambda f\over v}$, not $n={c\over \lambda f}$.
However, we could maybe think of it as effective wavelength and frequency, perhaps $\lambda_e$ and $f_e$ and get $n={c\over \lambda_e f_e}$.
Either way, let's use your version of the equation (maybe I made a mistake there), and we get:
${c\over v}={c\over f\lambda}=n=a+{b\over \lambda^2}$
Note that the right side is not proportional to $\lambda^{-2}$ because of the leading coefficient. Take out the middle and we get:
${c\over f\lambda}=a+{b\over \lambda^2}$
${c\over f}=a\lambda + {b\over \lambda}$
Extend the equation and we get:
${c\over f}=a\lambda+{b\over \lambda}+{c\over \lambda^3}+{d\over \lambda^5}+...$ $=\sum_{i=1}^{∞}{x_i\lambda^{3-2i}}$
Alternately, if we use my version, we get:
${\lambda f\over v}=a+{b\over \lambda^2}$
${f\over v}={a\over \lambda}+{b\over \lambda^3}$
Which extends to:
${f\over v}={a\over\lambda}+{b\over \lambda^3}+{c\over \lambda^5}+{d\over \lambda^7}+...$ $=\sum_{i=1}^{∞}{x_i\lambda^{1-2i}}$
In neither case do we get a mathematical contradiction, although I don't know enough about the physics to be certain there's no physical contradiction (though I somehow doubt such a contradiction would have gone unnoticed for this long).
Note the Wikipedia article says the Cauchy equation is only valid near the visible light spectrum (around 800-400 nm as I recall$^1$). It lists the first two coefficients for glass as ~1.5 and ~0.005 $\mu m^2$ (depending on the glass). So we need to graph things around x=0.4-0.8 for those coefficients to make sense. Here's the graph with the red line as $y=a$ and the blue line as $y={b\over x^2}$ -- notice how $a$ dominates at these wavelengths.
$^1$ It's actually 400-700 nm according to Wikipedia. Still, close enough.
Here are some links. I would put them in a comment, but it is a little long. No up or down votes needed.
Here is a derivation of the thin lens equation. https://byjus.com/physics/derivation-of-lens-maker-formula/
The RP Photonics Encyclopedia is always a good resource. Here is more information about focal length, with more links to thin and thick lenses and other topics. No derivations though. Focal Length
Here is similar from HyperPhysics. Lens Concepts
Also see Is there a more accurate form of the mirror equation 1f=1u+1v ?. It has information about lenses as well as mirrors.
Best Answer
I don't know what your teacher thought about while saying this, but this is not true. The power of the lens does depend on the refractive index $\mu_{lm}=\mu_l/\mu_m$ of the lens which is clear from the formula, you have given. $$P=\frac{1}{f}=(\mu_{lm}-1)\left( \frac{1}{R_1}-\frac{1}{R_2}\right)$$