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प्रश्न
Discuss the interference in thin films and obtain the equations for constructive and destructive interference for transmitted and reflected light.
उत्तर
- Interference in thin films:
- Let us consider a thin film of transparent material of refractive index p (not to confuse with an order of fringe n) and thickness d. A parallel beam of light is incident on the film at an angle i.
- The wave is divided into two parts at the upper surface, one is reflected and the other is refracted. The refracted part, which enters into the film, again gets divided at the lower surface into two parts; one is transmitted out of the film and the other is reflected back into the film.
Interference in thin films
- For transmitted light:
- The light transmitted may interfere to produce a resultant intensity. Let us consider the path difference between the two light waves transmitted from B and D. The two waves moved together and remained in phase up to B where splitting occurred.
- The extra path travelled by the wave transmitted from D is the path inside the film, BC + CD. If we approximate the incidence to be nearly normal (i = 0), then points B and D are very close to each other. The extra distance travelled by the wave is approximately twice thickness of the film, BC + CD = 2d. As this extra path is travelled in a medium of refractive index p, the optical path difference is,
δ = 2μd. - The condition for constructive interference in transmitted ray is,
2μd = nλ
Similarly, the condition for destructive interference in transmitted ray is
2μd = (2n-l) `λ/2`
- For reflected light:
- Wave while travelling in a rarer medium and getting reflected by a denser medium, undergoes a phase change of n or an additional path difference of `λ/2`.
- Let us consider the path difference between the light waves reflected by the upper surface at A and the other wave coming out at C after passing through the film.
- The additional path travelled by wave coming out from C is the path inside the film, AB + BC. For nearly normal incidence this distance could be approximated as, AB + BC = 2d. As this extra path is travelled in the medium of refractive index p, the optical path difference is, δ = 2μd.
- The condition for constructive interference for reflected ray is,
2μd + `λ/2` = nλ (or)
2μd = (2n – 1) `λ/2` - The additional path difference λ/2 is due to the phase change of n in rarer to denser reflection taking place at A. The condition for destructive interference for a reflected ray is,
2μd + `λ/2` = (2n + l)`λ/2` (or)
2μd = nλ
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संबंधित प्रश्न
Four light waves are represented by
(i) \[y = a_1 \sin \omega t\]
(ii) \[y = a_2 \sin \left( \omega t + \epsilon \right)\]
(iii) \[y = a_1 \sin 2\omega t\]
(iv) \[y = a_2 \sin 2\left( \omega t + \epsilon \right).\]
Interference fringes may be observed due to superposition of
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
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Answer in brief:
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