We have seen that thin film causes a path difference 2µt cos r between interfering rays. Now, when area of the film is small, the rays from various portions of the film reaching the eye have almost the same inclination. So it is the variation in thickness that gives fringes. Each fringe is the locus of all such points where thickness is same. Such fringes are known as Newton’s fringes or fringes of equal thickness. On the other hand when film has uniform thickness the path difference changes with r only. Each fringe in this case represents the locus of all points on the film, from which rays are equally inclined to the normal. Such fringes are called Haidinger’s fringes or fringes of equal inclination. To get such fringes the source must be an extended one.

**Newton’s rings: **It is a special case of interference in a film of variable thickness such as that formed between a plane glass plate and a convex lens in contact with it. When monochromatic light falls over it normally we get a central dark spot surrounded by alternatively bright and dark circular rings. When white light is used the rings would be coloured.

**Experimental Arrangement: **Let S be the extended source of light, rays from which after passing through a lens L falls upon a glass plate Gat 45°. After partial reflection these rays fall on a plano convex lens P placed on the glass plate E. The interference occurs between the rays reflected from the two surfaces of the air film and viewed through microscope M as shown Figure 4.6.

**Theory:** The air film formed is of wedge shape so the path difference produced will be

∆= 2 µt cos (r+θ) – λ/2

For normal inciden r = 0 so

∆= 2 µt-λ/2

At the point of contact t = 0

So µ=λ/2

The central fringe will be dark.

**Condition for maxima and minima**

2 µt = (2n + 1) λ/2

2 µt = nλ n = 0, 1, 2

we get alternatively bright and dark rings.

**Diameter of the rings:** From the Figure 4.7.

DB X DC = AD X DC

DB= DC= r, radius of ring under consideration.

CE = DA = BF = t, thickness of the film at B.

OG = OA = R, radius of curved surface of lens.

Then, r^{2}= t (2R- t) = 2Rt- t^{2} = 2Rt

t = r^{2}/2R

**Diameter of bright rings D _{n}** : From the condition of maxima

2 µ r^{2}n /2R = (2n + 1) λ/2 ,

r^{2}n = (2n + 1) λR / 2µ =( D_{n} / 2)^{2}

D _{n}^{2 } = 2 (2n + 1) λR / µ

= 2 (2n + 1) λR (For air, µ = 1)

D_{n} =√(2AR) √(2n + 1) = K √(2n + 1)

D_{n} = √odd number .

Thus, the diameter of bright rings are proportional to the square root of the odd natural numbers.

**Diameter of dark ring D _{n}:** Using the condition for minimum

2µ r_{n}^{2} /2R = nλ

which gives

(D_{n})^{2} = 4nλR

Or D_{n }= √(4 λR) √n

D_{n } = √n

Thus, the diameter of dark rings are proportional to the square root of natural numbers.

** **

**4.1 0.1 Measurement of Wavelength of light by Newton’s Rings**

For dark rings we know

(D_{n})^{2} /n = 4λR

λ = (D_{n})^{2} /n /4nR

But as the fringes around the centre are not very clear so n cannot be measured correctly.

To avoid any mistake one can consider two clear fringes nth and (n + p)th

So, (D_{n})^{2} /n = 4λR (D_{n+p})^{2} = 4(n+p) λR

Then ((D_{n+p})^{2} – (D_{n})^{2}) = 4(n+p) λR

Or λ = ((D_{n+p})^{2} – (D_{n})^{2}) / 4pR

Thus, measuring the diameters and knowing p and R, A can be measured.

**4.10.2 Measurement of Refractive Index of Liquid by Newton’s Rings**

For this purpose liquid film is formed between the lens and glass plate.

We have, as above,

which give

[(D_{n+p})

^{2}– (D

_{n})

^{2}]

_{liquid}= 4pλR/µ [(D

_{n+p})

^{2}– (D

_{n})

^{2}]

_{air}= 4pλR

One can see that rings contract with the introduction of liquid.