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Newton Rings

When a plano-convex lens of long focal length is placed on a plane glass plate, a thin film of air is enclosed between the lower surface of the lens and the upper surface of the plate. The thickness of the air film is very small at the point of contact and gradually increases from the centre outwards. The fringes produced with monochromatic light are circular. The fringes are concentric circles, uniform in thickness and with the point of contact as the centre. With monochromatic light, bright and dark circular fringes are produced in the air film.

S is a source of monochromatic light at the focus of the lens L₁. A horizontal beam of light falls in the glass plate B at 45˚. The glass plate B reflects a part of the incident light towards the air film enclosed by the lens L and the plane glass plate G. The reflected beam from the air film is viewed with the microscope. Interference takes place and dark and bright circular fringes are produced. This is due to the interference between the light reflected from the lower surface of the lens and the upper surface of the glass plate of G.

Theory: (i) Newton’s rings by reflected light: Suppose the radius of curvature of the lens is R and the air film of thickness t is at a distance of OQ = r, from the point of contact O.

Here, interference is due to reflected light. Therefore, for the bright ring

2 t cos θ = (2n – 1) λ/2              (i)

where, n = 1, 2, 3….etc.

Here, θ is small, therefore,    cos θ = 1

For air, = 1

2t = (2n – 1) λ/2                        (ii)

For the dark ring,

2 t cos θ = nλ

Or, 2t = nλ

where, n = 0, 1, 2, 3, …etc.           (iii)

EP × HE = OE × (2R – OE)

But, EP = HE = r, OE = PQ = t

and, 2R – t = 2R (approximately)

r² = 2R.t

or, t = r²/2R

Substituting the value of t in equation (ii) and (iii)

For bright rings

For dark rings,

R₂ = nλR                                    (vi)

R = √nλR                                   (vii)

When n = 0, the radius of the dark ring is zero and the radius of the bright ring is √(λR/R). Therefore, the centre is dark. Alternately dark and bright rings are produced.

Result: The radius of the dark ring is proportional to (i) √n (ii) √λ and (iii) √R. Similarly the radius of the bright ring is proportional to

If D is the diameter of the dark ring,

D = 2r =                         (viii)

For the central dark ring

N = 0

D = = 0

This corresponds to the centre of the Newton’s rings. While counting the order of the dark rings 1, 2, 3, etc. the central ring is not counted.

Therefore for the first dark ring.

N = 1

D₁ = 2√λR

For the second dark ring,

N = 2,

D₂ =

And for the nth dark ring,

Dn =

Take the case of 16th and 9th rings,