I had a question about finding the radius for bright Newton's rings. According to Wikipedia the final answer should be
$$ r = \sqrt{\lambda R\left(N+\frac{1}{2}\right)} $$
where $R$ is the radius of curvature and $N$ is a whole number starting from 0. When I was deriving the equation, I end up with:
$$ r = \sqrt{\lambda R\left(N+\frac{1}{2}\right) – d^2} $$
where $d$ is the distance from the edge of the circle to the ground. I looked it up, and this says that $d^2$ can be ignored because $d$ is small.
My question is, what is the basis for concluding that $d$ is small? Are there cases where this is not true?
Best Answer
In the derivation of the formula for the radius of a ring the intersecting chord theorem, Pythagoras , etc is used to show that $r^2= 2Rt -t^2 = t(2R-t)$ where $t$, your $d$, is the distance between the top of the bottom surface and the bottom of the lens surface.
The radius of curvature of the surface of the lens $R$ might be $1.00\, \rm m$ and $t$ would be a few wavelengths of light, say $10^{-6}\, \rm m $.
So $2R-t = 2\times 1.00 - 10^{-6} \approx 2.00$ to a very good approximation.
It is unlikely that the radius of curvature of the surface of the lens will be comparable to the wavelength of light.