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## Newtons Ring and Interference in thin Films, Physics tutorial

:Newton's RingsWhen the plano-convex lens with convex surface is placed on the plane glass sheet, air film of slowly increasing thickness outward is formed between lens and sheet. Thickness of film at point of contact is zero. If monochromatic light is permitted to fall normally on lens, and film is observed in reflected light, alternate bright and dark concentric rings are observed around point of contact. These rings were first found by Newton that is why they are known as Newton's rings.

Newton's rings are formed because of interference between light waves reflected from top and bottom surfaces of air film formed between lens and glass sheet.

Newton's rings experiment arrangementA lens is placed in contact with the plane sheet of glass as shown in figure given below. Len's lower surface is of very large radius of curvature. Due to this, the air film is formed between lower surface of lens and upper surface of plate. At the point of contact, thickness of the air-film is zero and it increases as one move away from this point of contact. When light rays reflected back and brought to the focus to the microscope, bright and dark fringes are achieved. Thus taking into account phase change at B, the n

^{th}bright and dark rings are therefore provided by2AB = (n - 1/2)λ For a bright ring.......................................Eq.1

and 2AB = nλ For a bright ring.......................................Eq.2

The interference pattern called as Newton's rings is achieved when light from the monochromatic source is reflected from the sheet of glass. Interference takes place between light reflected from lower surface of the lens and upper surface of the plane glass.

Radius of a Ring:Now refer to Figure given above to get the relation between radii of rings and wavelength of light. Here we will find out radius of either a dark or bright Newton's rings formed.

Let r

_{n}be the radius of n^{th}Newton's ring at A where film thickness is t= AB, and a is the radius of curvature of the lens surface of which A is a part.By theory of intersecting chords

(2a - t) x t = r

_{n}x r_{n}On expanding the above equation, we get

2at - t

^{2}= r^{2}_{n}On rearranging this equation, we get

2t = r

^{2}_{n}/a(As t is small as compared to a, thus t

^{2}is neglected)Condition for bright ring is

Thus r

^{2}_{n}/a = (n - 1/2)λ.......................................Eq.3Or r

^{2}_{n}= (2n - 1)λa/2 (Bright ring)The condition for dark ring is r

^{2}_{n}/a = nλ.......................................Eq.4= nλa (Dark ring)

These expressions can also be simply written in terms of diameter of ring. Assume D

_{n}is the diameter of the n^{th}ring, then Eq.3 and Eq.4 becomesD

_{n}= 2r_{n}= r_{n}= D_{n}/2Therefore D

^{2}_{n}= 2(2n - 1)λa (Bright ring)D

^{2}_{n}= 4nλa (Dark ring)Interference in thin Films:Thin-film interference takes place when incident light waves reflected by upper and lower boundaries of the thin film interfere with one another to create the new wave.

The thin film is very thin layer of medium concerned. Instances of thin film are the soap film or bubble and thin layer of oil spread over water surface. These layers are also cause interference patterns.

Let the thin transparent film of refractive index μ. A ray IA from the point monochromatic source is partially reflected as the ray AR and part refracted in material of film along AB, such that r is an angle of refraction. At point B, ray of light is partially reflected to A¢, and partially transmitted out. At A', ray will again get partially reflected along A'B' and refracted as ray A'R'. When these two rays, AR and A'R', meet, then interference takes place as shown in figure given below. It is to be noted that amplitude decreases from one ray to next.

Let A'D is the perpendicular to AR. Then optical path difference between these rays is

= μ(AB + BA') - AD...............Eq.5

But cos r = BT/AB = t/AB = t/cosr = BA' = AB

And sin i = Ad/AA' ...............Eq.6

Therefore AD = AA'sin i

AA' = AT + TA'

AA' = BT tanr + BT tanr

= 2t tanr

Therefore AD = 2t tanr sini

AD = 2t sin r/cos r sin i...............Eq.7

By using Snell's law as

μ = sin i/sin r = sin i = μ sin r...............Eq.8

Substitute Eq.8 in Eq.7, we get

AD = 2t sin r/cos r μ sin r

AD = 2μt. sin

^{2}r/cos rNow substitute value in Eq.5, for path difference, expression is

Path difference = μ(t/cos r + t/cos r) - 2μt sin

^{2}r/cosr= 2μt cos r [as sin

^{2}r + cos^{2}r = 1]At point A, ray is reflected when it is going from rarer to denser medium and suffers a path difference of λ/2 or a phase change of Π. But at B, reflection occurs when ray is going from denser to rarer medium, and therefore there is no phase change.

Therefore effective path difference between two rays is

2μt cos r - λ/2

Condition for destructive interference in film is provided by

2μt cos r - λ/2 = (2n - 1)λ/2

2μt cos r = nλ where n = 1, 2,.....

The condition for constructive interference is:

2μt cos r - λ/2 = nλ

2μt cos r = (2n + 1)λ/2

where n = 0, 1, 2,.....

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