  # Newton's rings This method for determining the wavelength of light was proposed by Sir Isaac Newton in his book Opticks, published in 1717. The experimental arrangement is shown in Figure 1.

A plano-convex lens of large radius of curvature R is placed on a plane glass plate with its curved surface downwards and is illuminated from above with a parallel beam of monochromatic light. Some of the light is reflected from the upper surface of the glass plate and some from the lower surface of the lens; interference thus occurs by division of amplitude, the fringes being localised in the air gap between the lens and plate.

At any point a distance r from the axis of the lens the path difference will be 2h, where h is the distance between the lens and the plate at that point (See Figure 2). The interference fringes are circular because the system is symmetrical about the centre of the lens. The radius of any ring is given by:

(2R - h)h = r2 so r2 = 2Rh – h2

But h2 is small compared with 2Rh and so: 2Rh = r2

The path difference (2h) is therefore r2/R

A phase change of π occurs when the light reflects from the top surface of the plate but not at the lower surface of the lens, and therefore:

For a bright ring viewed by reflection:   (2m + 1)λ/2 = rm2/R

For a dark ring viewed by reflection:      mλ = rm2/R

where m = 0, 1, 2, 3, etc and rm is the radius of the mth ring. If a graph is plotted of r2 against m for the dark rings a straight line should be produced with a gradient given by:

(rm2 - r12)/(m - 1) = λR

where r1 and rm are the radii of the first and mth rings respectively. (See Figure 3).

When doing the experiment it is much easier (and more accurate) to measure the diameter of the rings and then calculate their radius. A dark central spot should be obtained when viewed by reflection. The rings can be viewed by transmission by putting the microscope below the plate, and if this is done the equations for bright and dark rings should be interchanged as two phase changes will occur, producing an effective phase difference of 2π. A bright central spot should be obtained.

If white light is used a few coloured rings will be seen due to the different wavelengths of the different colours of light.

## Newton's rings and the refractive index of a liquid

Putting a liquid of refractive index n between the lens and the plate (Figure 5) will change the path difference to 2nh and give a formula for the m th dark ring of: mλ = [nrm2]/R

The radius of any given ring will be less with the liquid in place than without it.
This effect may be used to measure the refractive index of the liquid; the method is a good one since it is accurate and easy to perform, and only a small amount of the liquid is needed. Example problem

Calculate the ratio of the diameters of the fifth interference rings with and without water (refractive index 1.33) between the lens and the plate if the radius of curvature of lens is 0.50 m.

Without water:
rm2 = Rm= 0.5x5 xλ
With water:
rm2 = Rml/n = [0.5x5xλ]/1.33
Therefore:
ratio of radii = 1/(1.33)1/2 = 0.87

Example problem
A series of rings formed in Newton's rings experiment with sodium light was viewed by reflection. The diameter of the mth dark ring was found to be 0.28 cm and that of the (m + 10)th 0.68 cm. If the wavelength of sodium light is 589 nm, calculate the radius of curvature of the lens surface.
For the mth ring:      [(0.14x10-2)2 ]/R = mx589x10-9
and for the (m + 10)th ring:      [(0.34x10-2)2]/R= (m + 10)
where R is the radius of curvature of the lens surface.
Therefore:
0.342/0.142 = (m + 10)/m      giving m = 2.
Therefore, substituting for m, R = 1.66 metres.

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