What is the Newton Ring Experiment?
Calculate diameter for a Bright and Dark Ring. Why the central ring is Dark?
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 concentric-ring 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 plano-convex lens which a large radius of curvature has placed on the surface of the glass plate. The plano-convex lens meets at a contact C on a glass plate. The transparent glass plates held at 45 above to the plano-convex 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 plano-convex 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 air-to-glass 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.
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 plano-convex 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 = (2n-1)λ/2 (1)
For dark ring
In the Triangle (CAB); according to Pythagoras theorem
CB2 = AC2 + AB2
R2 = (R-t)2 + r2 (t2 <<< neglect)
r2 = 2Rt
t = r2/ 2R (3)
For diameter of bright ring; from equation (1)
2t + λ/2=nλ
Putting value of t from equation (3)
2 x r2/ 2R = (2n-1)λ/2
r2 = R x (2n-1)λ/2
the diameter is D=2r then
D2/4 = R x (2n-1)λ/2
So, the diameter of bright ring is directly proportional to the root of the (2n-1) odd natural number.
For diameter of Dark ring; from equation (2)
2r2/ 2R = nλ
r2 = nλR
D2 = 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 nth order of dark ring diameter is
Dn2 = 4nλR (6)
and next (n+p)th dark ring diameter is
D2n+p = 4(n+p)λR (7)
Subtract equation (6) and (7)
D2n+p – Dn2 = 4(n+p)λR – 4nλR
D2n+p – Dn2 = 4pλR
Problem 1: In Newton’s ring experiment, the diameter of 13th ring is 5.5 x 10-3m and diameter of 6th ring is 3.3 X 10-3m. If the radius of curvature of lens is 1m; calculate the wavelength of incident light.
Solution: Given that D13 = 5.5 x 10-3m; D6 = 3.3 X 10-3m; p = 13-6=7; R = 1m
l = D2n+p – Dn2/4pR
l = (5.5 x 10-3)2 – (3.3 x 10-3)2 / 4 X 7 X 1
l = 6900 A
Problem 2: In Newton’s ring experiment a source emit two types of light of wavelength is 6000 A and 5900 A. It is observed that the nth 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 nth dark ring.
Solution: Given that l1 = 6100 A =6.1 X 10-7m ; l2 = 5900 A0 =5.9 X 10-7m; R = 1m
The diameter of nth dark ring of l1 is
Dn2 (l1) = 4nRl1
The diameter of n+1th dark ring of l2 is
D2n+1 (l1) = 4(n+1)Rl2
According to question both the rings coincides than
4nRl1 = 4(n+1)Rl2
n+1/n = l1 / l2
1+1/n = 6.1 X 10-7 /5.9 X 10-7
1/n = 1.0339-1
n = 29
So the diameter of dark ring is
Dn2 (l1) = 4nRl1
Dn2 (l1) = 4 X 29 X 6.1 X 10-7 X 1
Dn (l1) = 0.0084 m