Aim :

To determine the wavelength of sodium light by measuring the diameters of Newton’s rings

Theory :

Fig. 6.1 shows the experimental setup of Newton’s ring. The formation of maximum intensities at some points and minimum intensities at the other due to the superposition of two coherent light waves (of same frequencies and constant phase difference) is called interference of light. The interference fringes are observed as an alternate pattern of bright and dark fringes. The interference at a point where the intensity of light is maximum, is called constructive interference (corresponds to bright fringe). For constructive interference, the two waves should have either same phase or a constant phase difference of

(6.1)

φ = 2nπ

where n = 0, 1, 2, .... Phase difference (φ) and path difference (δ) are related by the equation

(6.2)

φ = (2π/λ)δ

where λ is the wavelength of the incident light. So, for constructive interference the path difference between the light waves should be

(6.3)

δ = nλ

The interference at other point where the intensity of light is minimum, is called destructive interference (corresponds to dark fringe). For destructive interference, the two waves should have either same phase or a constant phase difference of

(6.4)

φ = (2n + 1)π

where n = 0, 1, 2, ... or a constant path difference of

(6.5)

δ = (2n + 1)λ/2

Interference fringes are obtained by dividing the single coherent source into two sources. This can be achieved by one of the following methods

1. by division of wave front, that is by taking (or considering) two secondary wavelets on the same wave front and superposing them

2. by division of amplitude, that is by separating the amplitude of single wave and reuniting them.

In the case of Newton’s ring interference is due to division of amplitude.

Figure 6.1: Experimental setup of Newton’s ring

When light is incident on a thin film (thickness of the order of wave length of the incident light), it suffers partial reflection and partial transmission at both upper as well as lower surfaces of the thin film. The transmitted light ray again suffers reflection at the lower surface. Interference occurs between the rays in the reflected and transmitted parts.

Similarly, in a wedge-shaped film, partial reflection as well as partial transmission also takes place. Moreover, the path difference changes from point to point which results into an interference fringe

Fig. 6.2 shows an wedge-shaped air film formed between the convex and plane glass plate inclined at an angle say θ. The refractive index of the film is μ. Ray AB is incident from a broad monochromatic source almost normally on the film. It suffers partial reflection (ray BE) and partial transmission (ray BC) on the convex surface. Ray BC again suffers partial reflection (ray CF) and partial transmission (not shown) on the plane surface at C.

The bright rings in the Newton’s ring, as shown in Fig. 6.3, are due to constructive interference between the reflected light rays BE and CF. The dark rings are caused by destructive interference between the same light rays BE and CF.

For a wedge-shaped thin film, the path difference between the rays BE and CF is given by

(6.6)

δ = 2μt cosα

Where t is the thickness of the film at B (or at D) and α is the angle of refraction at B. Since the angle of incidence is almost normal, we can assume cosα ~ 1. Note that here we ignore the reflections from top of the plano-convex lens and bottom of the plane circular glass plate since these reflections just contribute to the overall glare. The reflections of interest are only those involving the surfaces in contact.

Now by Stoke’s law, there is no phase change at the glass-air interface of the convex lens (because the wave is going from a higher to a lower refractive index medium), whereas the reflection at the air-glass interface of the plane glass plate undergoes an additional path difference of /2. Therefore, the net path difference is

(6.7)

δ + λ/2 = 2μt +λ/2

Since for bright fringe, net path difference is nλ we have,

(6.8)

2μt = (2n + 1)λ/2

Similarly, for dark fringe, the net path difference is (2n + 1)λ/2 and thus we have,

(6.9)

2μt = 2nλ

At the centre, no reflection occurs since the two glass surfaces are in intimate contact i.e. t = 0 or 2μt = 0. This is the condition for dark fringe. Hence the center of the pattern is always dark.

In the right-angled triangle OAB of Fig. 6.4,

OB^2 = OA^2 + AB^2

Or, R^2 = (R-t)^2 + rn^2

(6.10)

rn^2 = 2Rt

for t^2 << 2Rt

t = rn^2/2R

(6.11)

t = Dn^2/8R

where, rn = radius of nth ring,

Dn = diameter of the nth ring,

R = radius of curvature of the plano-convex lens

In practice, it is not possible to find the exact centre of the bull’s eye in order to obtain rn. Rather, the traveling microscope can measure an approximate diameter Dn for the interference ring. Therefore, substituting the value of t from Eq. 6.9, we obtain the diameters of the nth dark fringe as

(6.12)

Dn^2 = 4nRλ/μ

Since the human eye is more sensitive to small changes in low intensity, we will measure positions of dark fringes throughout the experiment.