Determination of refractive index of a liquid by Newton's rings

Theory of Newton's rings:

7)Determination of refractive index of a liquid by Newton's rings:

           First Newton's rings are ontaobta when there is an air film  between  the  glass  plate  and  the plano-convex lens. The system is placed in a metal container.  Measure  the  diameter  of  nth  and (n+p)the dark  ring  with  the  help  of  traveling microscope.

For air film:

Dₙ² = 4nλR

D²ₙ₊ₚ = 4(n+p)λR

.•. D²ₙ₊ₚ - Dₙ² = 4pλR    .....(1)

              Now pour the liquid whose refractive index is to be calculated in the container without disturbing the arrangement. The air film is now replaced by liquid. Now measure the diameter of nth and (n + p)th  dark  ring with  the  help  of traveling microscope.

For liquid film :

Let μ is the R.I. of the liquid

D'ₙ² = 4nλR
             μ

D'²ₙ₊ₚ = 4( n+p)λ.R
                      μ

.•. D'²ₙ₊ₚ - D'ₙ² = 4pλR      ......(2)
                               μ

Eq. (1) / eq. (2) gives

μ = D²ₙ₊ₚ - Dₙ²
      D'²ₙ₊ₚ - D'ₙ²                  .....(3)

           Eq. (3) gives the refractive index of the liquid by knowing the diameters of nth and ( n+p )th dark rings with and without liquid.

Newton's rings by with light (coloured rings)

Theory of Newton's rings:

5)Newton's rings by with light (coloured rings):

          Diameter of the Newton's ring depends on the wavelength of light used .  White  light  consists  of
 ( seven colours) continuous range of colours.  Each colour possesses a particular wavelength. So each colour produces its own system of rings. Only first , it cannot be seen.

           At a point, near the  point of  contact if  for a particular  wavelength  ( say violet colour ),  the condition of brightness will be satisfied,  then  it gives maximum intensity for that colour. At the same point, for other colours, the condition of darkness may be satisfied. Consequently that point will have average intensity due to red colour. The same is true for other colours.

6) Determination of wavelength of monochromatic light using Newton's rings:

              Let S be a monochromatic source of light which is at the focus of a lens L₁. So horizontal parallel rays fall on the glass plate B which is placed at 45° to the horizontal line. The glass plate B, partially reflects these parallel rays normally on the lens L placed on the glass plate G. Interference occurs between the rays reflected from the top and the bottom surfaces of the air film. Newton's rings formed can be viewed with the help of traveling microscope focussed on the air film. The position of traveling microscope is adjusted to get the centre of Newton's rings at the position of the intersection of the cross-wire of the eye piece. The microscope is moved until one of the cross wire becomes tangential to the 20th dark ring . This microscope reading is noted. Then the microscope is moved such that the cross wire becomes tangential to 15th, 10th, 5th dark rings respectively. Reading of microscope for each case should be noted . Microscope reading corresponding to the same rings on the other side of the centre should be noted. The reading are noted in the observation table:

Dₙ = Diameter of nth bright ring

Dₙ² =2(2n-1)λR             .....(1)

Dₙ₊ₚ= Diameter of (n+p)the bright ring

Dₙ₊ₚ²= 2[2(n+p)-1]λR   .....(2)

Dₙ₊ₚ² - Dₙ² = 4.pλ.R

λ = Dₙ₊ₚ² - Dₙ² 

              4pR                   .....(3)

              The radius of curvature of lens L can be found out by any simple method. Substituting the value of R and the mean value of P in equation (3), λ can be calculated.

Theory of Newton's rings

Theory of Newton's rings:-

3)Newton's rings with bright centre ( due to reflected light ).

If the plano-convex lens and glass plate are of different materials of refractive indices μ₁ and μ₂ and a liquid of refractive index μ
 ( lying between μ₁ and μ₂ ) is introduced between them. Then both the reflections from the film would be similar. So the centre of Newton's rings will be bright. This was shown by scientist young, using oil between crown and flint glass,
( μflint >μoil>μcrown ). If μ₁>μ>μ₂, then reflection from top and bottom surface of the liquid film takes place under similar conditions. In both the cases reflection takes place from the surface of optically rarer medium I (path difference=2t only )
Since at a point t=0; the condition of path difference=nλ =0 satisfies. So , central spot appears bright. If μ₁<μ<μ₂ then the central spot will also be bright. Here, also reflections from top and bottom surface of the liquid film takes place under similar conditions. In both the cases reflection takes place from the surface of optically denser medium. So additional phase change of π radius should be considered at interfering rays at the point of contact satisfies the condition for brightness. So the central spot appears bright.

4) Newton's rings by transmitted light

Newton's rings observed by transmitted light are exactly opposite to those observed by reflected light. In this transmitted light, the central spot is bright . By transmitted light:

The diameter of the dark ring,
Dₙ = √2(2n-1)λR

And the diameter for the bright ring,
Dₙ= 2√nλR

Theory of Newton's rings

Theory of Newton's rings:

2) Expression for the diameter of Newton's rings by reflected light:

                Let a plano-convex lens be placed on a plane glass plate. Let R is the radius of curvature of the lens and C is the centre of curvature. The thickness of the air film at point B is MO(=t) at a distance of ON=rₙ from the point of contact O.

                For interference due to reflected light,

The path difference between
two interfering rays  = 2μtcos( r+ϴ  ) +𝞴 /2

Here,     μ= 1 (for air film)
               r= 0 (for normal incidence)

              Geometrical Optics and Interference
And ϴ = 0 ( for large value of R )

.•. path difference=2t + 𝞴 /2  ...(1)

From figure A) ,
                 R² = rₙ²+(R-t)²
                 rₙ²= R²-(R-t)²
                 rₙ²= 2Rt-t²

                Since R >>t, t² becomes very small, so it can be neglected.

                 rₙ² = 2R.t
                 2t = rₙ²/R

Putting this value of 2t in eq.(1) we get,

Path difference= rₙ²/R+𝞴 /2    ....(2)

Where, rₙ = radius of the nth Newton's ring.

For dark rings:

The condition of dark ring is

Path difference = (2t+1)𝞴 /2

.•. rₙ²/R+𝞴 /2= (2n+1)𝞴 /2

.•. rₙ²/R=n𝞴

If Dₙ is the diameter of the nth Newton's rings then,

rₙ=Dₙ/2      ⇒    rₙ²=Dₙ²/4

.•. Dₙ²/4R =n𝞴  i.e. Dₙ²=4Rn𝞴

Dₙ = 2⎷nR𝞴

.•. Diameter of dark ring is the proportional to the square root of natural number
( Dₙ ∝ ⎷n)

For bright ring:

For condition for bright ring is that, path difference=n𝞴

.•. rₙ²/R + 𝞴/2=n𝞴

rₙ²/R= (2n-1)𝞴/2    where n= 1,2,3,......

Putting rₙ² = Dₙ²/4

.•. Dₙ²/4R = (2n-1) 𝞴/2

Dₙ² = 4R(2n-1) 𝞴/2

       =2𝞴R(2n-1)

Dₙ = ⎷2(2n-1)𝞴R

Diameter of the bright ring is proportional to the square root of odd natural numbers.

Dₙ ∝  (2n-1)

Newton's rings

Newton's rings:

           A plano-convex lens of large radius of curvature is placed with its convex surface on a glass plate. The air film is formed between the lower surface of the lens. AOB and upper surface of the glass plate G. The thickness of the film is zero at the point of contact O and gradually increases outward. If monochromatic light is allowed to fall normally on this film, and the film is viewed for reflected light then alternate bright and dark concentric rings are seen, with their centre as dark . These rings were first studied by  Newton and hence are known as Newton's rings. Newton's rings are formed as a result of interference between the light waves reflected from the upper and lower surface of the air film.

1) Formation of Newton's rings:

            Newton's rings are formed as a result of interference between the waves reflected from the top and bottom surfaces of the air film.

             When a ray of monochromatic light is incident normally on the air film, on reflection from the top and bottom surfaces of the film, the reflected rays interfere. The path difference between them ( ray 1 and 2) will be

               2𝜇.t.cos(r+ϴ ) + λ/2

Where   𝜇- Refractive index of the film.
                t - thickness at a point under                                consideration.
               r - angle of refraction at the upper                        face of the air.
               ϴ- angle of wedge.

               Now, for air film, 𝜇= 1; and for normal incidence r =0. And the large radius of curvature of plano-convex lens;ϴ becomes 0(nearly).

.•. path difference between rays 1&2=2t+ λ/2
                                                                      ...(1)

At the point of contact of the lens and glass plate, t = 0.

.•. from equation (1) path difference = λ/2

                 This is the condition for minimum intensity. Hence the centre spot of Newton's rings is dark

The condition for maximum intensity is path difference = nλ.

.•. from equation (1) 2t + λ/2 = nλ.

                  So for nth order, maximum intensity occur for a constant value of thickness t. In air film, thickness ‘t' remain constant along a circle.so brightness is in the form of circle. Different maximum intensity will occur for different values of ‘t'
 Similarly minimum intensity will also in the form of circular rings.
The maximum intensity circular rings occur when path difference is λ,2λ,3λ,....

                   The minimum intensity circular rings occur when path difference is λ/2, 3λ/2, 5λ/2,... i.e. circular rings of maximum and minimum intensities will occur alternatively. Each fringe ( or ring ) is a locus of constant film thickness and so these are the circular fringes of constant thickness.

Thin films

Thin films:-

               Thin films means a transparent layer of very small thickness in the range of 0.5 μm to 10 μm ; such as soap film or a layer of skin oil floating on water surface. When light is imcidein on a thin film, different colours are seen. This can be explained by considering interference between light reflected from the top and the bottom surface of a thin film. In case of thin films, interference occur due to i) reflected light and ii) transmitted light.

Interference in thin film of uniform thickness due to reflected light.

               Consider a thin transparent film of uniform thickness ‘t' and of refractive index ‘μ' be placed in air. Let a beam of light AB is incident on the upper surface l, at an angle' i as shown in figure A. At B, light suffers partial reflection along BE and partial refraction along BC. The angle of refraction is r. At C, it is again partially reflected along CD. At D it is partially refracted along DF.

               Incident beam AB suffers partial refraction at B  ( top surface )  and  at  C
 ( bottom surface). So we get a set of parallel beams  like  BE ,  DE.  etc.  These  beams originate from same incident beam. So these beams are coherent. When these reflected beams BE and DF superimpose, interference takes place.

               Let us calculate the path difference between the beams BE and BCDF. Let DM and BN be perpendicular to BE and CD respectively. Produce BH and DC to meet at G. Let t= BH is the thickness of the film.

By geometry ∠HGC = r ; ∠BDM =i ; ∠DBN= r
 .•.      BC = GC

The path difference between two reflected beams BE and DF=(BC+CD) in film-BM in air

          = μ ( BC + CD ) - BM
          = μ ( BC + CN + ND ) - BM ....(1)

We know that,

μ = sin i = BM / BD = BM
      sin r    ND / BD     ND

.•. BM = μ.ND

.•. Equation. (1) becomes, path difference
 
           = μ ( BC + CN + ND ) - μ. ND
           = μ ( BC + CN )
           = μ ( GC + CN )
           = μ . GN

In ∆ BGN,

cos r = GN = GN  ( •.• BG = BH+HG = t + t = 2t)
             BG     2t

.•. GN = 2.t cos r   .....(3)

Putting this value of GN in eq. (2) we get ,

Path difference =2.μ.t.cos r   .....(4)

               Eq. (4) does not represent a correct path  difference.  Because  whenever reflection takes place from the surface of optically denser medium a phase change of π radius ( or path difference λ/2) takes place.

                The film is optically denser than the  surrounding  medium  ( air ).
The ray BE originating  by reflection at the denser medium, suffers a phase change of  π radius due to reflection at B.

               ( No such change of phase occur for ray DF, as it is result of reflection at C, the surface of rarer medium ).

The correct path difference between reflected beams = 2μ.t.cos r + λ/2 = nλ

i) The firm appears bright if the path difference. = nλ

2μ.t.cos r + λ/2 = nλ

.•. 2μ.t.cos r = (2n - 1 ) λ/2

ii) The firm appears dark if path difference
     = (2n + 1) λ/2

.•. 2μ.t.cos r +  λ/2 = (2n + 1) λ/2


2μ.t.cos r  = nλ   where, n = 0,1,2,....

Interference of light

Interference of light:- 

                 Interference is the phenomenon exhibited by light when two or more light waves superimpose on each other.

                 The resultant light intensity at any point in the region of superposition is different from the intensities due to each wave. It means that the resultant intensity of light changes from point to point in the region of superposition. The modification of light intensity in the region of superposition is called interference of light.

                   In the region of superposition there are some points where the waves superimpose in such a way that the resultant intensity is greater than the sum of the intensities due to individual waves . This is constructive interference. There are some other points where the resultant intensity is less than the sum of the intensities due to individual waves. This is destructive interference.

a) condition for maximum and minimum intensity:

                   If two light waves having same speed  the phase difference between them. Constructive interference is said to take place only when the resultant light intensity is maximum. These points appear to be bright.

When phase difference 𝜹= 0, 2π, 4π, ...., 2nπ

i.e corresponding path difference = 0, λ,2λ,..,nλ.

                  So the condition for constructive interference is that the path difference between the two waves should be an integral multiple of the wavelength λ.

I.e. Path difference = nπ.   Where n = 0,1,2,....

                 Destructive interference is said to take place only when the resultant light intensity is minimum. So these points appear to be dark.

When phase difference,
 𝜹= π, 3π, 5π, ...,(2n+1)π

i.e Corresponding path difference,
=  λ,3λ,5λ,...,(2n+1) λ
    2  2   2                   2

                 If the amplitudes of the waves are not equal, there will not be complete destructive interference. The condition for destructive interference is that the path difference between the two waves should be an odd integral multiple of half the wavelength.

i.e. path difference= (2n+1) λ
where n=0,1,2,3,....

                Thus as the position of the point P on the screen changes, the path difference between the two waves at P changes and the intensity becomes alternatively maximum and minimum. This is called interference pattern.

b) Classification of interference phenomenon: 

Broadly interference phenomenon can be classified into two groups.

                The important condition for steady interference pattern is that the two light source must be coherent. It means the phase difference between them is always constant. Generally such type of light source can be obtained from a single and the same source of light by using different methods. There are two methods for of light by using different methods. There are two methods for producing such coherent sources. These are,

i) Division of wavefront and
ii) Division of amplitude.

                Depending on a particular method, the phenomenon of interference is studied. And so interference phenomenon can be studied under these two categories. i) Division of wavefront method and ii) Division of amplitude method.

i) Division of wavefront method : 

                In this method, the wavefront generated by a particular source ( known as a incident wavefront ) is divided into two separate wavefronts by using some techniques as in case of superimpose on each other gives a stedy interference pattern. Young's experiment , Fresnel's biprism experiment are the examples of interference by division of wavefront.

ii) Division of amplitude method : 

                  In this method, the amplitude of incident wave is divided into two beams by using partial reflection and refraction. ( As in case of Michelson's experiment). Then by using some techniques these two beams superimpose on each other gives a stedy interference pattern. Michelson's interferometer, interference in thin films, Newton's ring  experiment are the examples of interference by division of amplitudes.