newton s rings n.
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  2. INTRODUCTIONThe formation of Newton’s rings is an important application of interference of light wave from the opposite faces of a thin film of variable thickness.

  3. FORMATION OF NEWTON’S RINGS A thin air film of increasing thickness in all direction from one point can be easily obtained by placing a plano-convex lens of large radius of curvature on a plane glass plate.

  4. When the air film is illuminated normally by monochromatic light the observed interference fringes are circular concentric rings with centre as the point of contact. The rings are circular since these are the loci of points of equal optical film thickness. The rings gradually become narrower as their radii increase. • AB is a wave incident at point B on plano convex lens. BD and B1D1 are waves originating by reflection from lower surface of plano convex lens and upper surface of glass plate. These two interfere to give interference circular fringes by reflected light. CE and C1E1 are transmitted waves on the other side of film. These also produce interference circular fringes by transmitted light.


  6. RADIUS OF NEWTON’S DARK RINGSrn=sqrt (nλR/µ)………..(1)

  7. RADIUS OF NEWTON’S BRIGHT RINGS rn=sqrt[{(n+1/2)λR}/µ ]…….(2)

  8. Diameter of nth ring is given by Dn=2rn From eqn. (1), D1=2sqrt(λR/µ) D2=2sqrt(2λR/µ) D3=2sqrt(3λR/µ) The difference in the diameters of two nearest dark rings is given by : D2-D1 =2sqrt(λR/µ)[sqrt2-1]=0.414*2sqrt(λR/µ) D3 – D2 = 2sqrt(λR / µ)[sqrt3-sqrt2]=0.318*2sqrt(λR / µ) D4 -D3 = 2sqrt(λR/µ)[sqrt4-sqrt3]=0.268* 2sqrt(λR / µ) Hence the rings gradually becomes narrower as their radii increases . It can also be shown that the bright rings also gradually become narrower as their radii increases.

  9. NEWTONS RING BY TRANSMITTED LIGHT Condition of maxima: 2µt=nλ Condition of minima: 2µt=(n+1/2)λ

  10. Difference between newton’s rings by reflected light and transmitted light. In the reflected system, there is a phase change of П between the two reflected light rays while in the transmitted system there is a phase change of 2П between the two transmitted light rays . Therefore the condition of maxima and minima are opposite in two systems. Hence the centre in the reflected system is dark while in the transmitted system is bright.

  11. DETERMINATION OF WAVELENGTH BY METHOD OF NEWTON’S RINGS From eqn.(1), diameter of nth dark ring is Dn=2sqrt(nλR) Dn2=4nλR From eqn.(1), diameter of (n+m)thdark ring is Dn+m2 = 4 (n+m) λR D n+m 2-Dn2= 4mλR λ=(D n+m 2-Dn2 )/4mR ………….(3) Hence measuring , n,m, Dn , Dn+m and R , λ can be determined experimentally.

  12. EXPERIMENTAL SET UP The cross wire of the microscope is fixed at a particular dark ring, say, nth and reading is noted on microscope. The microscope is moved to diametrically opposite side and the cross wire is fixed on the same ring and reading is noted. The difference between the two readings will be the diameter of that particular ring(nth). This procedure is repeated for another ring ,say,(n+m)th . Thus diameter of (n+m)th ring is obtained The radius of curvature R of the lower surface of the lens is measured by a spherometer and by putting the values in equation (3), wavelength of light is determined.

  13. THANKS