Newton's rings in a special case of wedge shaped film in which an air film is formed between a glass plate and a convex surface of lens. The thickness of air film is zero at the center and increases gradually towards the outside.
When a plano-convex lens of large focal length is placed on a plane glass plate, a thin air film is formed between the lower surface of plano-convex lens and upper surface of glass plate. When a monochromatic light falls on this film the light reflected from upper and lower surfaces of air film, and after interference of these rays, we get an inner dark spot surrounded by alternate bright and dark rings called Newton's rings. These rings are first observed by Newton and hence called Newton's rings.
Experimental Arrangement for Reflected Light
The experimental arrangement for Newton's rings experiment. A beam of monochromatic source S is made parallel by using a convex lens L. The parallel beam of light falls on a partially polished glass plate inclined at an angle of o45. The light falls on glass plate is partially reflected and partially transmitted. The reflected light normally falls on the plano-convex lens placed on plane glass plate.
This light reflected from upper and lower surface of the air film form between plane glass plate and plano-convex lens. These rays interfere and rings are observed in the field of view. The figure shows the reflection of light form upper and lower surfaces of air film which are responsible for interference.
Formation of Bright and Dark Rings
The air film can be considered as a special case of wedge shaped film. In this case, angle wedge is the angle made between the glass plate and tangent drawn to curved surface of plano convex lens.
The path difference between two interfering rays reflected by air film is
Δ = 2μt Cos (r + Θ) − λ⁄2 ...... (1)
where μ is the refractive index of the air film, t is the thickness of air film at the point of reflection (say point P) is angle of refraction and Θ is angle of wedge.
In this case the light normally falls on the plane convex lens for the angle of refraction r = 0. Further, as we use a lens of large focal length the angle of wedge Θ is very small. So
Cos (r + Θ) = Cos Θ = Cos 0 = 1
and thus the path difference
Δ = 2μt − λ⁄2 ...... (2)
At point of contact t = 0, therefore,
Δ = λ⁄2
Which is the condition of minima. Hence at the centre or at point of contact there is a dark spot.
Case (i) Bright Rings
The condition for bright rings is path difference Δ = nλ therefore
Δ = 2μt − λ⁄2 = nλ where n = 0, 1, 2, 3.........
2μt = (2n+1)⁄2 λ
2μt = (2n−1)⁄2 λ ...... (3)
Where n = 1, 2, 3.
Case (ii) Dark Rings
In case of dark rings, the path difference,
Δ = (2n−1)⁄2 λ
Where n = 1, 2, 3............
Therefore Δ = 2μt − λ⁄2 = (2n−1)⁄2 λ
2μt = nλ ...... (4)
Thus corresponding to n = 1, 2, 3...... we observe first, second third..... etc. bright or dark rings. In Newton's rings experiment the locus of points of constant thickness is a circle therefore the fringes are circular rings.
Diameter of Bright and Dark Rings
The plano-convex lens BOPF is place on glass plate G and O is the point of contact. Suppose, C is the centre of the sphere OBPF from which the plano-convex lens is constructed. P is point on the air film of thickness of air film is t at point P. The light is incident and reflected form the upper and lower surface of air film, and rings are formed. AP is the radius of ring passes through point P. According to property of circle
AP × AB = AO × AL
r2 = t × (2R − t) :: AL = OL − OA
Where R is the radius of curvature of lens.
r2 = 2Rt − t2
Since R is very large and t is very small, we can write
r2 = 2Rt or t = r2⁄2R
Substituting this value of t in equation (..), we get,
2μ r2⁄2R = (2n−1)⁄2 λ
r2 = (2n−1)⁄2 λR⁄μ
This expression contains n, i.e., r is a function of n. Thus it is better to use rn in place of r. If Dn is the diameter of nth bright ring then we have r = rn = Dn/2 and can write
Dn2⁄4 = (2n−1)λR⁄2μ
Dn2 = 2(2n−1)λR⁄μ ...... (5)
Where n = 1, 2, 3........ Similarly for dark rings
2μt = nλ or 2μ r2⁄2R = nλ or r2 = nλR⁄μ
If Dn is diameter of nth dark ring then
Dn2⁄4 = nλR⁄μ
Dn2 = 4nλR⁄μ ...... (6)
Where n = 1, 2, 3.........