Question

Discuss the phenomenon of interference in thin films. Obtain the condition for Maxima and Minima.​

Answer 1

Answer:

The condition for observing : Minima — If the path difference is a multiple of half the wavelength. Maxima — If path difference is an integral multiple of the wavelength thus waves interfere constructively

Explanation:

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Answer 2

Answer:

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

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=tcosr(1)

Now,

BD = (2t) tan r (2)

BG = BD sin i

BG = (2t) tan r sin i

BG=2tμsinr(sinrcosr) [μ=sinisinr]

BG=2μtsin2rcosr(3)

Substituting (i) and (iii) in Δ :

Δ=μ(tcosr+tcosr)−2μtsin2rcosr

=2μtcosr(1−sin2r)

Δ=2μtcosr

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 ±λ2 to path difference

Δ=2μtcosr±λ2

For Destructive Interference:

Δ=nλ

2μtcosr±λ2=nλ

2μtcosr=(2n±1)λ2(n=0,1,2)

This is the required expression for constructive Interference or Maxima.

For Destructive interference:

Δ=(2n±1)λ2

2μtcosr±λ2=nλ

2μtcosr=nλ

This is the required expression for destructive interference.