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Newton’s Rings

Netwon’s Rings are a phenomenon of wave optics and occur if there is a small path difference of coherent light.
They occur for example for large spheres on a glass plate with a very small gap between sphere and plate.

Some light is reflected from the glass sphere’s surface and some is reflected from the glass plate’s surface.
Because the light is coherent, we get constructive and destructive interference.

For the gap being 1/4 Lambda, the path in the gap and back is lambda/2, which results in destructive interference. Same for path lengths in the gap of 3/2 lambda, 5/2, etc. Say, for gaps that are odd multiples of lambda 1/4.
For even multiples of lambda/4, say lambda/2 , 3/2 Lambda etc, we get constructive interference.

Say, we simply can count the bright rings. For 666nm red light for example, three rings are 1000nm = 1 micrometer, which is the length of the air gap.

Newton’s rings are used in lens production, together with interferometers to find how close the current shape is from the wanted shape.

Thus surface accuracies are given in “(Newton’s) Rings”. The less rings, the better the shape got to be (and the more expensive the lens will be). Also a diametre should be given, for example five rings in 18mm diameter.

According to ISO DIN 10110 the default specs are: Wavelength 546,07 nm, 5 Rings at at 10 mm clear aperture

Below is a version of Newton Rings occuring between a sphere under test and a reference sphere

Newtonian Image Equation

newton-linsengleichung

Image formation of a lens. The values z und z’ are marked red.

The Newtonian equation is an equation of ray optics named after the English physicist Isaac Newton.

It reads
f^2 = z \cdot z'

with

z = ObjectDistance - FocalLength

and

z'= ImageDistance - FocalLength

Solved for ObjectDistance we get:

ObjectDistance = FocalLength + \frac {f^2}{ImageDistance - FocalLength}

Solved for ImageDistance we get:

ImageDistance = FocalLength + \frac {f^2}{ObjectDistance - FocalLength}

This Newtonian Image equation is often used instead of the “lens equation”. Here z is the differences between object distance and focal length and z ‘ is the difference between image distance and focal length.

Be an object point 20mm left of the first Focal point of a lens of 30mm focal length.
Then it’s image is -\frac{30 * 30}{-20} = 45mm to the right of the second Focal point (“-” 20 because the left of the focal point).
Be an object point 10mm left of the first Focal point of a lens of 30mm focal length.
Then it’s image is -\frac{30 * 30}{-10} = 90mm to the right of the second focal point (“-” 10 because the left of the focal point).

The advantage of Newton’s equation is that you can determine the focal point and distances from these focal points of lenses relatively easy, while principal points can be relatively difficult to determine.