I am familiar with the concept of optical path, constructive and destructive interference. The basic premise to discuss these concepts is coherence, which is why I am perplexed by the phenomenon of Newton's rings.

From what I understand, this experiment was first conducted by Robert Hooke, with non-coherent white light (from a candle I think?).

If the waves that enter the lens are of arbitrary phase, and of all possible wavelengths, why should there be a diffraction pattern?


Two salient reasons:

  1. The interference is between the reflexions from two neighboring surfaces;
  2. The distance between these two surfaces is small - only a few wavelengths of visible light.

Point 1 means that it is only the phase difference that is important in determining the throughgoing / reflected light for a monochromatic wave. So, all monochromatic waves, whatever their random phase, produce the same interference pattern.

But we are dealing with all wavelengths. So let's look at point 2.

Near the center of the pattern, the distance between the two interfering surfaces can be approximated by a quadratic dependence on the distance $r$ from the center of the pattern. Therefore, the throughgoing intensity as a function of radius for light of wavelength $\lambda$ is proportional to:

$$\frac{1}{2}\left|1-\exp\left(i\,\frac{\pi\,r^2}{R\,\lambda}\right)\right|^2 = \sin\left(\frac{\pi\,r^2}{2\,R\,\lambda}\right)^2$$

where $R$ is the radius of curvature of the curved surface, if the interference pattern is formed by bringing a convex lens into contact with a glass flat, for example. Therefore the nulls in the pattern happen at radiusses $0,\,\sqrt{4\,R\,\lambda},\,\sqrt{8\,R\,\lambda},\,\cdots$.

Now we add the effect of all wavelengths. We can only see a narrow band of wavelengths, so we are looking at the sum of the interference patterns with nulls at radiusses $0,\,\sqrt{4\,R\,\lambda},\,\sqrt{8\,R\,\lambda},\,\cdots$ for $\lambda$ varying between $400{\rm nm}$ and $750{\rm nm}$. This means that the first null happens at a range of radiusses that varies only over a range of about $\pm 20\%$ - the "smear"width is well less than the distance between the first and second null. So, even with the spread over visible wavelengths, the first nulls line up pretty well. The second null less well and so forth. You see a series of colored nulls - the coloring is because different wavelengths have their nulls at different positions, but the nulls are still well enough aligned to see their structure. As you move further from the center, the nulls become more tightly packed and the accuracy of the alignment for all visible wavelengths becomes coarser than the null spacing, which means that we can no longer see fringes. This is exactly what happens in Newton's rings - the fringe visibility fades swiftly with increasing distance from the center.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.