I have performed experiments in my college laboratory on Newton's rings to find radius the of curvature of the convex lens used. I always get a dark center. Is it possible to get a bright center? If yes, how?
[Physics] Newton’s rings experiment
diffractionlensesoptics
Related Solutions
Not necessarily. The lens equation, as provided by Nordic, is
You can play around with the numbers, and a few things should become obvious.
1) If you increase the radius of one surface to a very large number, the lens becomes either a plano-convex or a plano-concave lens. For the moment, think only about a plano-convex lens, with R2 essentially infinite. Then
1/F = (n-1)(1/R1)
or F = R1 / (n-1)
where n is the refractive index.
2) So, for n < 2, the focal length of a plano-convex thin lens will be greater than the radius of curvature, and for n > 2, the focal length will be less.
Most optical materials have an index of refraction in the range of 1.3 to 1.7 for visible light, so for most lenses, the focal length will be greater than the radius of curvature. Diamond, though, has an index of refraction of about 2.4, so such a lens will reverse the usual order. And although it is reflective in the visible, germanium is transparent in the mid to far infrared, and has an index of refraction of about 4 at these wavelengths.
Intuitively, the radius of curvature has to depend on the index of refraction of the glass. If the index were $1$, the lens would have no effect at all. If the index were very high, say $10$, it would not take much curvature to get a given f'ocal length. Clearly we cannot just say the radius of curvature is twice the focal length.
You have applied the lens maker's formula correctly to your problem. There is no problem with $R=f$, in fact that is always true for $\mu_2=1.5\mu_1$. We are assuming a thin lens in this formula, so the diameter of the lens must be small compared with $R$. If the focal length is $30$ cm and the diameter of the lens is $1$ cm the thickness is twice the height of a circular segment. Given $R=30, c=1$ we have $h=R-\sqrt{R^2-(\frac c4)^2}=30-\sqrt{900-\frac 14}\approx 0.004$ so the lens is about $1$ mm thick
Best Answer
One can do this but it is difficult and you would need a great deal of patience and optical experimentation skill.
The reason that the centre is almost always dark is that the classic Newton's rings experiment simply involves putting the convex lens in contact with the reference optical flat. The lens touches the flat at the centre; therefore near the centre the two reflexions from the surface of the convex lens and from the flat are almost in phase. They are also opposite in sign, since one is from light going from glass to air (convex lens into the space), the other is from light going from air to glass.
What you would need to do for a non-dark centre, and indeed what you should do if you do not want to marr the convex lens's surface by the test, is to hold the two surfaces a half wavelength apart (at the centre). To do this, you would need to mount the flat or lens in a translation stage and bring the two surfaces together very carefully until the ring pattern can be seen. If you can do this successfully, you should be able to adjust for a bright centre.