What is the Newton Ring Experiment?
Calculate diameter for a Bright and Dark Ring. Why the central ring is Dark?
Answer:
Newton Ring Experiment
When viewed with monochromatic light, Newton’s rings appear as a series of concentric, alternating bright and dark rings (Constructive and destructive interference) centered at the point of contact between the two surfaces. When we take white light source, it forms a concentricring pattern of rainbow colors, because the different wavelengths of light interfere at different thicknesses of the air layer between the surfaces. This concentric bright and dark ring first observed by Newton’s and known as Newton ring phenomenon; “in which an interference pattern is created by the reflection of light between two surfaces—a spherical surface and an adjacent flat surface”.
Newton Ring Experiment arrangement:
The Newton ring experiment arrangement shown in the figure. The planoconvex lens which a large radius of curvature has placed on the surface of the glass plate. The planoconvex lens meets at a contact C on a glass plate. The transparent glass plates held at 45^{0} above to the planoconvex lens; a monochromatic light source emits light on a transparent glass plate. This glass plate reflects the light on the lens and glass plate. A part of the incident light is reflected by the curved surface of the lens and a part is transmitted which is reflected back from the plane surface of the plate. These two reflected rays interfere and show an interference pattern. This interference pattern observed with the help of a microscope which placed vertically toward the lens.
Formation of Newton Ring
The ring is formed when light is reflected from a planoconvex lens of a long focal length placed in contact with a plane glass plate. A thin air film is present between the plate and the lens. The thickness of the air film present between the lens and plates varies from zero at the point of contact to some value t. If the lens plate system is illuminated with monochromatic light falling on it normally, concentric bright and dark interference rings are observed in reflected light.
According to figure, a ray AB incident normally on the system gets partially reflected at the bottom curved surface of the lens (Ray 1) and part of the transmitted ray is partially reflected (Ray 2) from the top surface of the plane glass plate. The rays 1 and 2 are derived from the same incident ray by division of amplitude and therefore are coherent. Ray 2 undergoes a phase change of p upon reflection since it is reflected from airtoglass boundary.
If the ray interfere there are two constructive and destructive interference present in interference pattern i.e. for normal incidence cos r = 1 and for air film µ = 1.
Constructive interference
2t = (2m+1)λ/2
 Destructive interference
2t = mλ
Why the central ring is Dark?
At the point of contact of the lens with the glass plate the thickness of the air film is very small compared to the wavelength of light therefore the path difference introduced between the interfering waves is zero. As a result, the interfering waves at the center are opposite in phase and interfere destructively; so that central point is dark.
Circular fringes with equal thickness: The locus of points having the same thickness fall on a circle having its centre at the point of contact, the fringes are circular; because each maximum or minimum is a locus of constant film thickness.
Calculation of diameter for a Bright and Dark Ring
A planoconvex lens LL’ placed on a glass plate. Let the radius of curvature of lens is R and the r is the radius of Newton’s ring; t is the thickness of the air film. The path difference is defined as
2µtcosq + λ/2=nλ
If µ = 1 for air film and q=0 for normal incidence
2t + λ/2=nλ
For bright ring
2t = (2n1)λ/2 (1)
For dark ring
2t=nλ (2)
In the Triangle (CAB); according to Pythagoras theorem
CB^{2} = AC^{2} + AB^{2}
R^{2} = (Rt)^{2} + r^{2 }(t^{2} <<< neglect)
r^{2} = 2Rt
t = r^{2}/ 2R (3)

For diameter of bright ring; from equation (1)
2t + λ/2=nλ
Putting value of t from equation (3)
2 x r^{2}/ 2R = (2n1)λ/2
r^{2} = R x (2n1)λ/2
the diameter is D=2r then
D^{2}/4 = R x (2n1)λ/2
So, the diameter of bright ring is directly proportional to the root of the (2n1) odd natural number.

For diameter of Dark ring; from equation (2)
2t=nλ
2r^{2}/ 2R = nλ
r^{2} = nλR
D^{2} = 4nλR
So the diameter of the dark ring is directly proportional to the root of the natural number.
Expression for the wavelength of used source:
If the n^{th }order of dark ring diameter is
D_{n}^{2} = 4nλR (6)
and next (n+p)^{th }dark ring diameter is
D^{2}_{n+p} = 4(n+p)λR (7)
Subtract equation (6) and (7)
D^{2}_{n+p} – D_{n}^{2} = 4(n+p)λR – 4nλR
D^{2}_{n+p} – D_{n}^{2} = 4pλR
Problem 1: In Newton’s ring experiment, the diameter of 13^{th} ring is 5.5 x 10^{3}m and diameter of 6^{th} ring is 3.3 X 10^{3}m. If the radius of curvature of lens is 1m; calculate the wavelength of incident light.
Solution: Given that D_{13} = 5.5 x 10^{3}m; D_{6 }= 3.3 X 10^{3}m; p = 136=7; R = 1m
l = D^{2}_{n+p} – D_{n}^{2}/4pR
l = (5.5 x 10^{3})^{2 }– (3.3 x 10^{3})^{2} / 4 X 7 X 1
l = 6900 A^{0}
Problem 2: In Newton’s ring experiment a source emit two types of light of wavelength is 6000 A^{0} and 5900 A^{0}. It is observed that the n^{th} dark ring due to one wavelength coincides with n+1 dark ring due to other. If the radius of curvature of plano convex lens is 1 m than calculate the diameter of n^{th }dark ring.
Solution: Given that l_{1} = 6100 A^{0} =6.1 X 10^{7}m ; l_{2} = 5900 A^{0 }=5.9 X 10^{7}m; R = 1m
The diameter of n^{th }dark ring of l_{1} is
D_{n}^{2 }(l_{1}) = 4nRl_{1}
The diameter of n+1^{th }dark ring of l_{2} is
D^{2}_{n+1} (l_{1}) = 4(n+1)Rl_{2}
According to question both the rings coincides than
4nRl_{1} = 4(n+1)Rl_{2}
n+1/n = l_{1 }/ l_{2}
1+1/n = 6.1 X 10^{7} /5.9 X 10^{7}
1/n = 1.03391
n = 29
So the diameter of dark ring is
D_{n}^{2 }(l_{1}) = 4nRl_{1}
D_{n}^{2 }(l_{1}) = 4 X 29 X 6.1 X 10^{7} X 1
D_{n} (l_{1}) = 0.0084 m
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