Question

Compare , about the fringes between michelson's and newtons ring exp

Answer 1

21/21 MAHE – MIT – BE – ENGINEERINGPHYSICSINTERFERENCE OF LIGHTNewton’s ringsi)Rings are of constant thickness (i.e, Fizeau’s fringes)ii)Order of the rings increases outwards.iii)Radii are proportional to square root of natural numbers (for dark rings) or natural odd numbers (for bright ringsiv)Centre is always dark in the reflected systemMichelson’s ringsi)Rings are of constant inclination (Haidinger’s fringes)ii)Order of the rings decreases outwardsiii)Angular radii proportional to square rootof ring number (not the order).Countedfrom the center.iv)Either dark or bright depending on the relative positions of the mirrors.Applications: 1. Determination of wavelength:The Fact that whenever the movable mirror moves by 2, a fringeoriginates or vanishes at the center is used to determine from the equation2d = N , where d is the distance moved and N, the number of fringesoriginated or vanished.2.Determination of refractive index:When a thin film (whose refractive index n is to be determined) ofthickness t is introduced on the path of one of the interfering beams, anadditional path difference (nt – t)2 = 2t(n-1) will be introduced. As aresult there will be shift of fringes. If m fringes shift, then, 2t(n-1) = mfrom which n can be determined.3.Measurement of small difference in wavelength:This is based on the fact that, each wavelength will produce its ownfringe system. By moving the mirror M1, the two fringe systems can bemade to be in step (Consonance) or out of step (dissonance)If d is the distance, through which M1must be moved between successivepositions of consonance or dissonance, then2d = m= (m-1) (+), from which we get, d22A