Hints and formulae :
* from triangle 1 , BM / BE = sin i
BM = BE sin i ----------------> 1
* According to Snell’s Law , µ = sin i / sin r ,
Sin i = µ sin r ----------------> 2
* from triangle 2 , BL / BD = sin r ( BL = 1/2 BE )
1/2 BE / BD = sin r
BE = 2 BD sin r ---------------> 3
* from triangle 2 , DL / BD = cos r (DL = t)
t / BD = cos r ----------------> 4
Interference due to the Reflected Beam
Derivations :
from E , draw a normal EM to BC. The path difference between the waves BC and EF ,
δ = (BD+DE)M - (BM)A
Since, µ = λM / λA , i.e., λM = µ λA
δ = µ(BD+DE)A - (BM)A (from the diagram,BD=DE)
So, δ = µ(2BD) - (BM)
From equation 1, BM = BE sin i
So, δ = 2µBD - BE sin i
From equation 2, Sin i = µ sin r
So, δ = 2µBD - BE(µ sinr)
From equation3, BE = 2 BD sin r
So, δ = 2µBD - 2µBDsin2r
δ = 2µBD (1-sin2r) : δ = 2µBD cos2r
From equation 4, t / BD = cos r
δ = 2µBD cosr . cosr
δ = 2µBD(t/BD).cosr
δ = 2µt cosr
* Any light ray undergoing Reflection at the surface of a Denser medium will suffer a Phase change of Π or an additional Path difference of λ/2.
* At B due to reflection, there is an additional Path difference λ/2
So the effective path difference is , δ =2µt cosr + λ/2
For constructive Interference ,δ = nλ
i.e., δ =2µt cosr + λ/2 = nλ (or) 2µt cosr + λ/2 = nλ
So, 2µt cosr = λ/2 (2n-1)
For destructive Interference , δ =(2n+1)λ/2
i.e., δ =2µt cosr + λ/2 = (2n+1)λ/2
(or)
2µt cosr + λ/2 = (2n+1)λ/2
So, 2µt cosr = nλ
If light is incident Normally ,i = 0 and hence r = 0 ,
therefore the condition for Bright fringe is ,
2µt = (2n-1)λ/2
And the condition for Dark fringe is ,
2µt = nλ
