To produce Newton’s rings a plano-convex lens of large radius of curvature is placed on a glass plate so that its convex surface faces the plate, a thin air film of progressively increasing thickness in all directions from the point of contact between the lens and the glass plate is very easily formed
To produce Newton’s rings a plano-convex lens of large radius of curvature is placed on a glass plate so that its convex surface faces the plate, a thin air film of progressively increasing thickness in all directions from the point of contact between the lens and the glass plate is very easily formed (Fig. 22 a). The air film thus possesses a radial symmetry about the point of contact. When it is illuminated normally, preferably with monochromatic light, an interference pattern consisting of a series of alternate dark and bright circular rings (Fig. 22 b), concentric with the point of contact is observed.
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(a) (b) Fig 22 Theory of Newton’s rings: Let us consider a ray of
monochromatic light AB incident at the point B on the upper surface of the air
film (Fig. 23). One portion of the ray is
reflected from point B and goes upwards along BR. The other part refracts into
the air film along BC. At point C, part of the light is again reflected along
CB1R1. The two reflected rays along BR and BCB1R1
are derived from the same source and are coherent. They will produce
constructive and destructive interference depending on their path difference.
Let t be the thickness of the film at the point B1 and let the
tangent to the convex surface be inclined at an angle θ with the horizontal. Let us draw B1D
and B1E perpendiculars to BD and BC respectively. Now the optical path
difference between the reflected rays BD and BCB1 is expressed by
[ ∆ = µ (BC + CB1) – BD
[ µ=(Original distance/Aparent distance Fig 23 ∆ = µ (BE + EC + CB1) – BD ---------------------------(i) where µ is the refractive index of the film with respect to air. Now, <DB1B = i = angle of incident and <BB1E = r = angle of refraction. [<DBB1 + <BDB1 + <DB1B = 1800 ; or, 900
- i + 900 + <DB1B = 1800
From equ. (i) ∆ = µ (BE + EC + CB1)
– µ BE = µ (EC + CB1) --------------------(ii) Let BN and CN be the normal to the upper and
lower surfaces of the film respectively. Hence we get, <CNB = θ and <PCB =
<CBN + <CNB = r + θ, and <PCB1 = <PCB = r + θ In the Fig. 23, B1M is perpendicular from B1
on the lower surface of the air film and the extension of BC intersects it at
O. Thus we have the relation <B1OC = <PCB = r + θ Also <CB1O = <PCB1= r + θ = <B1OC. Thus B1OC is an isosceles
triangle. Hence CB1 = CO Equ. (ii), therefore, reduces to ∆ = µ (EC + CO) = µ EO = µ B1O cos (r + θ) [∴ If t is the thickness of the film at the
point B1 and B1M = MO = t.
We still have to consider a
path difference of Then the total optical path difference
between the two rays, ∆ = 2µt cos (r + θ) ± For constructive interference, 2µt cos (r + θ) ± λ/2 2µt cos (r + θ) = (2n -1)λ/2 The minus sign has been selected since n
cannot have a value of zero for bright fringes seen in reflected light. For destructive interference, 2µt cos (r + θ) ± 2µt cos (r + θ) = nλ ------------------(iv) dark In practice, a thin lens of extremely small curvature is used and hence the angle θ becomes negligible. Furthermore, the experiment is so designed that the light is incident almost normally on the film and and therefore cos r = 1. Accordingly Eqns. (iii) and (iv) reduce to 2µt = (2n -1)λ/2 --------------------------(v) bright and 2µt = nλ --------------------------(vi) dark Let
us now compute the radius of any ring. Let R be the radius of
curvature of the convex surface as shown in the Fig. 24.
From the right angled triangle OFB1, we can get the relation R2 = rn2
+ (R - t)2 rn2
= 2Rt – t2 where rn is
the radius of the nth circular ring corresponding to the constant
film thickness t. Here t is extremely small that t2 may be neglected
compared to 2Rt. Then
Substituting the value
of t in equs. (v) and (vi), we have Fig 24 rn2
= (2n -1)λR/(2µ) rn2
=nλR/µ Now the square of the diameters of the bright and dark rings are, therefore, given by
Determination of wavelength:
(a) (b)
Fig. 25 Let us consider the nth order ring (Fig. 25 b). For an
air film µ = 1. Therefore, the
square of the diameter
of the nth order bright or dark ring will be
D_n^2=4nλR ---------------------------(xii) dark Now we have to take another ring, p ring from this nth ring onwards. The diameter of this (n+p)th ring is also measured. Thus we have D_(n+p)^2=2[2(n+p)-1]λR Now applying {(xiii) - (xi)} and {(xiv) - (xii)}; we will get for both the dark and the bright rings the relation D_(n+p)^2-D_n^2=4pλR or, Eqn. (xv) is used to compute the wavelength λ of the monochromatic light
that employed to illuminate the film. Determination of the radius of curvature of the lens: The eqn. (xv) can be rearranged as
Thus, if the wavelength λ of the light used
is known then the eqn. (xvi) can be used to determine the radius of curvature
of the surface of the lens in contact with the plane glass plate. Problem 2: In a Newton’s ring experiment, the diameter of the 4th and the 8th rings are 3.21 mm and 4.55 mm respectively. If the radius of curvature of the surface of the convex lens in contact with the plane glass plate is 105.76 cm then what will be the wavelength of light used in this experiment?
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