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newton’s rings experiment with animation

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Newton’s Ring:

Newton’s rings is a process in which Circular bright and dark fringes obtained due to air film enclosed between a Plano convex lens and a glass plate.

An air wedge film can be formed by placing a plano-convex lens on a flat glass plate.The thickness of film is zero where the lens and the plate are in contact with each other.The thickness of the film gradually increases in all directions as we go away from the centre. If such a film is illuminated by a parallel beam of monochromatic light from the top and also viewed from the same direction,dark and bright consecutive circles will obtained.These rings are called Newton’s rings.

These rings are the result of constructive and destructive interference produced by thin film in the region between eyes and lens.

experimental arrangement for newton's ring

Experimental arrangement for newton’s rings experiment:

The experimental setup for Newton’s rings is shown in figure.

Light beam from source “S’ fall on convex lens from where the rays becomes parallel and strikes the glass plate ‘G’.The glass plate ‘G’ partly reflected beam and partly refract toward the mirror.The mirror ‘M’ reflect back the beam towards plano convex lens.The thickness of air film between glass plate ‘P’ and plano convex increases.Fro centre ‘o’ to outwards .So at point of contact the path difference becomes ‘λ/2’ and destructive interference take place.Similarly when there is constructive interference the bright circular rings is seen.As a result a series of dark or bright rings is seen,these are called newton’s rings.

Size of Newton’s rings:

 

image of newton's ring experiment

At the point of contact there will be dark circle due to 180 change in phase of the light ray reflected from lower surface of the air film.

The radii of bright and dark circles can be found by considering the figure.In this figure consider a circle of radius r which is formed at thickness “t” of air film.If “R” be the radius of curvature of the lens then in the triangle DBO1:

\ <span style="color: #ff0000;">{ (BD) }^{ 2 }\quad +\quad { ({ O }_{ I }D) }^{ 2 }={ (B{ O }_{ 1 }) }^{ 2 }\quad \quad \quad .......(bY\quad Pathagorous\quad theorem)\\ { (r) }^{ 2 }\quad +\quad { (R-b) }^{ 2 }={ (R) }^{ 2 }\\ { r }^{ 2 }={ R }^{ 2 }\quad -{ (R-b) }^{ 2 }\\ { r }^{ 2 }={ R }^{ 2 }\quad { - }{ R }^{ 2 }-{ b }^{ 2 }+2Rb\\ As\quad { b }^{ 2 }\quad is\quad very\quad small\quad as\quad compared\quad to\quad R,therefore\quad it\quad can\quad be\quad neglected.\\ { r }^{ 2 }=2Rb\quad \quad .............(1)\\ For\quad Maximum:\\ The\quad circle\quad will\quad be\quad bright\quad if\quad the\quad path\quad difference\quad between\quad the\quad interfering\\ rays\quad is\quad odd\quad integral\quad multiple\quad of\quad \frac { \lambda  }{ 2 } \quad i.e\\ 2nb=(m+\frac { 1 }{ 2 } )\quad \lambda \\ For\quad n=1\\ 2b=(m+\frac { 1 }{ 2 } )\quad \lambda \\ For\quad First\quad bright\quad ring\quad m=0\\ { 2b }1=(0+\frac { 1 }{ 2 } )\quad \lambda =\frac { 1 }{ 2 } \quad \lambda \\ For\quad second\quad bright\quad ring\quad m=1\\ { 2b }_{ 2 }=(1+\frac { 1 }{ 2 } )\quad \lambda =\frac { 3 }{ 2 } \lambda \\ For\quad 3rd\quad bright\quad ring\quad m=2\\ { 3b }_{ 3 }=(2+\frac { 1 }{ 2 } )\quad \lambda =\frac { 5 }{ 2 } \lambda \\ Similarly\quad for\quad Nth\quad bright\quad ring\quad m=N-1\\ { 2b }_{ N }=[(N-1)+\frac { 1 }{ 2 } ]\quad \lambda =(N-\frac { 1 }{ 2 } )\lambda \\ Substituting\quad { 2b }_{ N }\quad in\quad equation\quad (1)\quad we\quad get\\ { r }_{ N }^{ 2 }=(N-\frac { 1 }{ 2 } )\lambda R\\ { r }_{ N }=\sqrt { R(N-\frac { 1 }{ 2 } )\lambda  } \quad \quad \quad \quad \quad .....(2)\\ \\ Where\quad N=1,2,3....for\quad 1st,2nd,3rd,....bright\quad rings\quad respectively.\\ For\quad Minimum:\\ \\ The\quad circle\quad will\quad be\quad dark\quad if\quad the\quad path\quad difference\quad between\quad interfering\quad rays\quad is\\ integral\quad multiple\quad of\quad \lambda ,the\quad wavelength\quad of\quad light,i.e\\ 2nb=m\lambda \\ for\quad air,refractive\quad index\quad =n=1\\ 2b=m\lambda \quad \quad \quad \quad \quad ....(3)\\ Put\quad value\quad of\quad 2b\quad from\quad equation\quad (3)\quad in\quad (1)\\ { r }^{ 2 }=m\lambda R\\ r=\sqrt { m\lambda R } \quad \quad .....(4)\\ Where\quad m=1,2,3,......\quad for\quad \quad zeroth,1st,2nd,3rd,\quad \quad \quad \quad dark\quad rings\quad respectively.\\  </span>

Why newton’s rings are circular in nature?

Newton’s rings are circular in nature because the air film is symmetrical about the point of contact between the plano convex lens and the glass plate.

Newton’s ring experiment with animation:

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