**1 Answer**

written 6.6 years ago by | • modified 5.5 years ago |

Consider a thin transparent film of refractive index μ.Let the bounding surface be strictly parallel to each other so that thickness is uniform all over the film. Consider a monochromatic beam of light AB incident at the upper bounding surface, it is partially reflected in the direction BC and refracted in the direction BF. Successive reflections and refractions give two reflected rays BC and DE derived from one original ray AB. But they follow different paths and their phase will be different. Ray BC starts from B but DE has to travel BF +FD in a medium of refractive index μ. To find the path difference, drop the perpendicular from D to BC.

Therefore optical path difference Δ= μ(BF +FD)-BH ……………..(1)

In triangle BGF, cosr = t/BF

So, BF= t/cosr

Similarly, in triangle DGF, FD =t/cosr.

In triangle BHD sin i = BH/BD

```
Sin i = BH/2BG
BH = 2BG sin i
```

In triangle BGF, tan r = BG/GF

```
BG = GF tan r
BG = t tanr
```

Substituting BF, FD BH in equation (1)

Δ = μ(t/cosr + t/cosr ) -2t tanr sin i

Sini = μ sinr

Δ = 2μt/cosr- 2μtsinr/cosr sinr

Δ =2μt(1-$sin^2$ r)/cosr

Δ = 2μt$cos^2$ r/cosr

Δ = 2μtcosr.

But since reflection at point ‘B’ takes place at the surface of denser medium ,path difference of λ/2 should be taken into account.

Therefore actual path difference is Δ = 2μtcosr ±λ/2…………………….(2)

Condition for constructive interference (or maxima or brightness)]

If the OPD is an integral multiple of λ, then the waves interfere constructively.

Δ =nλ

From equation (2)

2μtcosr±λ/2 =nλ

2μtcosr =(2n±1)λ/2…………………………………(3)

Condition for destructive interference (or minima or darkness)

If OPD is odd multiple of λ/2, then the rays interfere destructively,

Δ =(2n±1)λ/2

From equation (2)

2μtcosr ±λ/2 =(2n± 1)λ/2

2μtcosr =nλ-----------------------------------------------------(4)

In reflected system intensity of maxima is 14% and minima is zero so the contrast between dark and bright fringe is good but in transmitted system intensity of maxima is 100% and minima is 85% so the contrast between dark and bright is poor. So, visibility of fringes is more in reflected light than the transmitted system.