**2 Answers**

written 5.4 years ago by |

Consider a thin film of uniform thickness (t) and R.I (μ)

On Reflected side,

The ray of light R1 and R2 will interfere.

The path difference between R1 and R2 is,

Δ = μ(BC + CD) − BG

BC = CD = t/cosr..........(1)

Now,

BD = (2t) tan r .......(2)

BM = BD sin i

BM = (2t) tan r sin i

BM = 2tμsinr(sinr / cosr)

BM = 2μt(sin2r/cosr)..........(3)

Substituting (i) and (iii) in Δ :

Δ = μ(t / cosr + t / cosr)−2μt(sin2r / cosr)

= 2μtcosr(1−sin2r)

Δ = 2μtcosr

For transmitted system :

The transmitted rays CT1 and ET2 are also derived from the same incident ray AB and hence they are coherent.

Path difference = △ = μ(CD + DE) – CL

For constructive interference :

2μtcos r = nλ

For destructive interference :

2μtcos r = (2n – 1)λ/2

written 7.8 years ago by |

Consider a thin film of uniform thickness (t) and R.I (μ)

On Reflected side,

The ray of light BF and DE will interfere. The path difference between BF and DE is,

$$Δ = μ (BC + CD) - BG $$

$$BC = CD = \frac{t}{cos r}..........(1)$$

Now,

BD = (2t) tan r .......(2)

BG = BD sin i

BG = (2t) tan r sin i

$BG = 2t μ sin r(\frac{sinr}{cosr})$ $[μ = \frac{sin i}{sin r}]$

$BG = 2 μt \frac{sin^2r}{cos r}......(3)$

Substituting (i) and (iii) in Δ :

$$Δ = μ (\frac{t}{cos r} + \frac{t}{cos r}) - 2 μt \frac{sin^2r}{cosr} $$

$$ = \frac{2μt}{cosr}(1 - sin^2 r)$$

$$Δ = 2μt cos r$$

This is a geometric path difference. However, there is a phase change of π, as ray BF is reflected from a denser medium. Hence we need to add $±\frac{λ}{2}$ to path difference

$$Δ = 2μt cos r ± \frac{λ}{2} $$

**For Destructive Interference:**

$$Δ = nλ$$

$$2μt cos r ± \frac{λ}{2} = nλ $$

$$2μt cos r = (2n ± 1) \frac{λ}{2}.....(n = 0,1,2,...)$$

This is the required expression for constructive Interference or Maxima.

**For Destructive interference:**

$$Δ = (2n ± 1) \frac{λ}{2}$$

$$2μt cos r ± \frac{λ}{2} = nλ$$

$$2μt cos r = nλ$$

This is the required expression for destructive interference.