optics4

Chapter III

OPTICS

Lecture 3.4

Books:(1) Optics, 3

edition: Ajoy Ghatak, McGraw-Hill

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Interference in Wedge shaped film (non-parallel films):

A wedge shaped film is the one in which surfaces are not parallel but makes some angle say α as shown in the figure.

In fig. We show the surface GH and G₁H₁ inclined at

Angle α .

Note that the thick-ness of the film is not constant but vary from

point to point.

Let µ be the refractive index of the thin film.

In fig let AB is the incident ray and BR and BC be the reflected and refracted rays respectively at point B.

The ray BC is reflected at point C. So CD is the reflected ray and

DR₁ is the emergent ray.

The interference pattern will be observed due to superposition Of BR and DR₁ rays.

Draw normal DE and DF on BC and BR respectively.

The path difference between these rays is

( ---(

Nnnnnnnnnnnnnnnnnnnnnnnnn ---(2)

S

---(3)

Now draw a normal DS on surface GH and extend it till it meet

extended BC at some point say P

In right angle triangle CSD and CSP,

DS = SP = d, CD = CP,

angle CSD = angle CSP = 900 ----(4) Thus

Angle CPD = angle CDP = r + α ---(5) Using (4), we get from eqn (3)

---(6)

---(6)

In right angle triangle DEP, we have

Cos (r + α) = EP/DP

EP = DP Cos(r+α) = 2d Cos (r+α) ---(7)

Thus From eqn (6) we get

---(8)

---(9)

---(10)

Spacing between consecutive dark or bright fringes:

---(12)

Let the nth maximum is formed at distance X_n from the edge.

From figure we observe that

---(15)

Using eqn (15), eqn. (12) becomes,

For (n+1)th maximum we have,

2X_n+1tan α cos (r+α) = ---(17)

Taking difference of eqn (16) and (17) we shall get that the fringe width,

---(18)

---(19)

For nearly normally incidence i.e. r = 0 and small angle i.e. Sin α = α, Finge width becomes

---(20)

2(X_n+1 - X_n ) tan α cos (r+α)

Newton Rings:

_{When a plano-convex lens with its convex surface is placed on a} plane glass plate, an air film of gradually increasing thickness is formed between the two.

_{The thickness of the film at the point of contact is zero. If}

monochromatic light is allowed to fall normally, and the film is

Experimental setup:

The experimental arrangement of obtaining Newton's rings is shown in figure. L is a plano convex lens

of large radius of curvature. This lens with its convex surface is placed on a plane

glass plate G.

The lens makes contact with the plate at

O. Light from an extended monochromatic source such as sodium.

lamp falls on a glass plate G' held at an angle 45° with the horizontal.

The glass plate G' reflects a part

of the incident light towards the air film enclosed by the lens L and the glass plate G.

A part of the incident light is reflected by the curved surface of the lens L and a part is transmitted which is reflected

back from the plane surface of the plate.

These two reflected rays interfere and give rise to an interference pattern in the form of circular rings.

Newton’s rings are formed due to

interference between the waves reflected from the top and bottom

surfaces of the air film formed between the plates. The formation of Newton’s rings can be explained

with the help of the Fig.

AB is a monochromatic ray of light, which

falls on the system. A part is reflected at B (glass-air boundary) which goes out in the form of ray R₁ without any phase reversal.

The other part is refracted along BC. At point C it is again reflected and goes out in the form of ray R₂

The reflected rays R₁ and R₂ are in a position to produce interference fringes as they have been derived from the same ray AB and hence fulfill the condition of interference.

As the rings are observed in the reflected light, the path difference between them is

(2μt cos r + λ/2). ---(1)

For air film μ = 1 and for normal incidence r = 0. Hence in this case, path difference is

(2 t + λ/2). ---(2)

At the point of contact t = 0, and the path difference is λ/2, which is the condition of minimum intensity. Thus the central spot is

For nth maximum, we have

2 t + λ /2 = nλ ---(3)

This expression shows that a maximum of a particular order will occur for a constant value of t.

In this system, ‘t’ remains constant along a circle. Thus the maximum is in the form of a circle. For

different value of ‘t’, different maxima will occur. Hence we get a number of concentric bright circular

rings.

In a similar way, this can be shown that minima are also in the circular form.

2 t + λ /2 = (2n+1)λ/2

Diameters of Newton Rings:

Let LOL' be the lens placed on a glass

plate G. The curved surface LOL' is the part of spherical surface with centre at C.

Let R be the radius

of curvature and r be the radius of nth bright ring corresponding to the constant film of

thickness t.

As

Discussed earlier for bright fringes we have ,

2 t + λ/2 = nλ ---(5)

We can write above eqn as,

2t = (2n – 1) λ/2 ---(6)

We now use the property of the circle

EP × PF = PO × PQ ---(7)

Substituting the required values in above eqn we get

r × r = t × (2R – t) = 2Rt – t2 ~ 2Rt (approximately) ---(8).

r2 = 2 R t or t = r2/2 R.---(9)

Thus for a bright ring (using eqn (9) in eqn (6)),

2r2/2 R = (2n – 1) λ /2

Replacing r by D/2, we get the diameter of nth bright ring as D_n2 /4= (2n −1)λR/2

or D_n2 = 2(2n – 1)λR ---(11)

or D_n = ---(12)2R (2n -1)

Thus the diameters of the bright rings are proportional to the square roots of odd natural numbers as (2n –1) is an odd

number.

Similarly for a dark ring (use eqn (9) in eqn (4)) or 2r2/2R = nλ

or D_n2 = 4 nλR (using r = D/2)

or D_n = 2 sqrt(nλR) sqrt (n) ---(13)∝

Thus diameters of dark rings are proportional to the square roots of natural numbers.

If Dm and Dn are the diameters of the mth and the nth rings we have

D_m2 _{− D}

n 2= 4(m − n)λR ---(14)

We can find the wavelength using eqn (15),

---(16)

To find the refractive index of a liquid, it is introduced

between the lens and glass plate and the experiment is repeated as before.