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Reflection by thin layers. In Fig. 35-42, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays r1and r2interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35- 2 refers to the indexes of refraction n1, n2andn3, the type of interference, the thin-layer thickness Lin nanometres, and the wavelength λin nanometres of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where Lis missing, give the second least thickness or the third least thickness as indicated.

Short Answer

Expert verified

The thickness of the thin layer is 329nm.

Step by step solution

01

Given data

  • The refractive index of first medium is n1=1.50.
  • The refractive index of the thin film is n2=1.34
  • The refractive index of the third medium n3=1.42
  • The maximum intensity occurs at wavelength λ587=587nm.
02

Interference in thin films

Bright colours reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colours can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

2L=m+12λn2m=0,1,2,... (Maxima—bright film in the air)

Where λis the wavelength of the light in air, Lis its thickness, and n2is the film’s refractiveindex.

03

Determining the 2nd least thickness for this arrangement of materials.

Here, in this case, light travels in medium with n1=1.50and incident on the film whose refractive index is n2=1.34.And then, the light gets reflected of the third surface n3=1.42while travelling through the film.As a result, the condition for constructive interference or maximum intensity is

2L=m+12λ587n2 (Constructive)

The 2nd least thickness which m=1is

role="math" localid="1663149600515" 2L=1+12587nm1.34L=3587nm41.34=329nm

Hence the thickness of the thin layer is 329nm.

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Most popular questions from this chapter

In two experiments, light is to be sent along the two paths shown in Fig. 35-35 by reflecting it from the various flat surfaces shown. In the first experiment, rays 1 and2 are initially in phase and have a wavelength of 620.0nm. In the second experiment, rays 1 and2 are initially in phase and have a wavelength of 496.0nm . What least value of distance L is required such that the 620.0nmwaves emerge from the region exactly in phase but the 496.0nmwaves emerge exactly out of phase?

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