S A B C N P E T AIR t i r r i r r L L M i Q O P r U U
Here i and r are angle of incidence and angle of Refraction. We drop a perpendicular AE from the point A on The lower surface LL and extend AE and CB to A point P where they meet. So clearly <EAB = <ABO = <OBC = r Where O is the foot of the perpendicular drawn from point B on AC.
The difference in path between two rays AT and CQ can be calculated. So first draw CN normal to AT and AM normal to BC. Now <ABE = /2 – r And <PBE = -[( /2 –r)+r+r] = /2 - r Here <APC = r <EAB = <ABO = <OBC = r
………….(1) This path difference for reflected light is not correct enough because on the basis of electro- magnetic theory when light is reflected from the surface of optically denser medium (air-medium interface) a phase change equivalent to a path difference /2 occurs.
The correct path difference here will be ……………..(2)
(1)If path difference x=n, where n = 0,1,2,3… constructive interference will take place and bright films will appear. So condition for maxima will be from equ(2)
If path difference x= (2n+1) /2 where n = 0,1,2,3.. destructive interference takes place and films appear dark.
Here interference pattern will not be perfect Because intensities AT and CQ will not be the same And their amplitude are different. Amplitude depends on amount of light reflected and transmitted through the films. Intensity never vanishes completely and perfectly dark fringes will not be observed. But for multiple reflection intensity of minima will be zero.
a at att ar 1 atr atr 2 atr 3 atr 4 atr 5 atr 6 atr 7 atr 8 atr 2 t atr 4 t atr 6 t atrt atr 3 t atr 5 t atr 7 t 123 4 5 rarer denser rarer
Amplitude of incident ray a Reflection coefficient = r and r 1 Transmission coefficient from rarer to denser medium = t Transmission coefficient from denser to rarer medium = t The amplitudes of the reflected rays are ar, atrt, atr 3 t, atr 5 t, atr 7 t……
When ray 1 is reflected from the surface of denser medium it undergoes a phase change Rays 2, 3, 4… are all in phase but out of phase with ray 1 by Resultant amplitude of 2, 3, 4, 5.. is A = atrt+atr 3 t+atr 5 t+atr 7 t+….. = attr[1+r 2 +r 4 +r 6 +……] = attr
According to the principle of reversibility tt = 1- r 2 So the resultant amplitude of 2,3,4,… is equal in Magnitude of the amplitude of ray 1 but out of Phase with it.