1. Newton’s Rings
Another method for observing interference in light waves is to place a planoconvex lens on top of a flat glass surface, as in Figure 24.8a
the air film between the glass surfaces varies in thickness from zero at the point of contact to some value t at P.

2. If the radius of curvature R of the lens is much greater than the distance r, and if the system is viewed from above light of wavelength λ, a pattern of light and dark rings is observed (Fig. 24.8b).
The interference is due to the combination of ray 1, reflected from the plate, with ray 2, reflected from the lower surface of the lens. Ray 1 undergoes a phase change of 180° on reflection, because it is reflected from a boundary leading into a medium of higher refractive index, whereas ray 2 undergoes no phase change, because it is reflected from a medium of lower refractive index.
Rewrite the equations 9,10 where the
thickness of the air change from 0 to d
For contractive interference nair=1
For destructive interference where
m=0,1,2,......

3. The contact point at O is dark, because there is no path difference and the total phase change is due only to the 180° phase change upon reflection. Using the geometry shown in Figure 24.8a, we can obtain expressions for the radii of the bright and dark bands in terms of the radius of curvature R, and vacuum wavelength λ. From the figure we get
Where r the distance from O we want to get the air thickness at P
Using the Binomial Series when r<<R
Then , and
But d for bright ring
Equating the two equations we get
And for the dark rings well have radii of

4. How to solve a thin –film interference
1. Identify the thin film causing the interference, and the indices of refraction in the film and in the media on either side of it.
2. Determine the number of phase reversals: zero, one, or two.
3. Consult the following table, which contains Equations 24.9 and 24.10, and select the correct column for the problem in question.

5. EXAMPLE:
Calculate the minimum thickness of a soap-bubble film (n = 1.33) that will result in constructive interference in the reflected light if the film is illuminated by light with wavelength 602 nm in free space.
Solution:
Solve 2nt= λ/2 for the thickness t, and substitute:

6. EXAMPLE:
Semiconductors such as silicon are used to fabricate solar cells—devices that generate electric energy when exposed to sunlight. Solar cells are often coated with a transparent thin film, such as silicon monoxide (SiO; n = 1.45) to minimize reflective losses (Fig. 24.9). A silicon solar cell (n = 3.50) is coated with a thin film of silicon monoxide for this purpose. Assuming normal incidence, determine the minimum thickness of the film that will produce the least reflection at a wavelength of 552 nm.
Solution: Solve 2nt= λ/2 for t, the required thickness:

7. EXAMPLE
A pair of glass slides 10.0 cm long and with n = 1.52 are separated on one end by a hair, forming a triangular wedge of air as illustrated in Figure When coherent light from a helium–neon laser with wavelength 633 nm is incident on the film from above, 15.0 dark fringes per centimetre are observed. How thick is the hair?
Solution
Strategy The interference pattern is created by the thin film of air having variable thickness. The pattern is a series of alternating bright and dark parallel bands. A dark band corresponds to destructive interference, and there is one phase reversal, so 2nt=mλ should be used. We can also use the similar triangles in Figure to obtain the relation t/x =D/L. We can find the thickness for any m, and if the position x can also be found, this last equation gives the diameter of the hair, D.

9. The Michelson Interferometer
A. A. Michelson (1852–1931), splits a light beam into two parts and then recombines the parts to form an interference pattern.
The device can be used to measure wavelengths or other lengths with great precision

10. -A ray of light from a monochromatic source is split into two rays by mirror M0 , which is inclined at 45° to the incident light beam. -Mirror M0, called a beam splitter, transmits half the light incident on it and reflects the rest. One ray is reflected from M0 vertically upward toward mirror M1, and the second ray is transmitted horizontally through M0 toward mirror M2. -Hence, the two rays travel separate paths L1 and L2. After reflecting from M1 and M2, the two rays eventually recombine at M0 to produce an interference pattern, which can be viewed through a telescope.

11. The interference condition for the two rays is determined by their path length differences.
As M1 is moved, the fringe pattern collapses or expands, depending on the direction in which M1 is moved.
For example, if a dark circle appears at the centre of the target pattern (corresponding to destructive interference) and M1 is then moved a distance λ/4 toward M0, the path difference changes by λ/2.
The wavelength of light is then measured by counting the number of fringe shifts for a given displacement of M1. If the wavelength is accurately known, mirror displacements can be measured to within a fraction of the wavelength.

12. the distance between the two mirrors d1and the number of fringes is m1, the relation is
The relation between the number (m1-m2) of fringes and the distance (d1-d2) between the two positions of the mirrors is given by

13. USING INTERFERENCE TO READ CD’S AND DVD’S
The data on these(CD’s and digital video disks DVD’s) disks are stored digitally as a series of zeros and ones, and these zeros and ones are read by laser light reflected from the disk.
Strong reflections (constructive interference) from the disk are chosen to represent zeros and weak reflections (destructive interference) represent ones.

14. As the disk rotates, the laser beam reflects off the sequence of bumps and lower areas into a photodetector, which converts the fluctuating reflected light intensity into an electrical string of zeros and ones.
Example
Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced mm, causing the interferometer pattern to reproduce itself 1700 times. Determine the wavelength of the light. What colour is it?
Solution:

15. Example:
Light of 587.5nm is used in a Michelson interferometer. When the movable mirror is moved, 1780 change from dark to dark patters are observed. How far was the mirror moved?