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

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,n2and n3, the type of interference, the thin-layer thickness in nanometres, and the wavelength λ in nanometres of the light as measured in air. Where is missing, give the wavelength that is in the visible range. Where is missing, give the second least thickness or the third least thickness as indicated.

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, n2and n3, 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.

In Fig. 35-45, two microscope slides touch at one end and are separated at the other end. When light of wavelength 500 nm shines vertically down on the slides, an overhead observer sees an interference pattern on the slides with the dark fringes separated by 1.2 mm. What is the angle between the slides?

Three electromagnetic waves travel through a certain point P along an x-axis. They are polarized parallel to a y-axis, with the following variations in their amplitudes. Find their resultant at P.

E1=(10.00μV/m)sin[2×1014t]E2=(5.00μV/m)sin[2×1014t+45°]E3=(5.00μV/m)sin[2×1014t-45°]

In Fig. 35-32a, a beam of light in material 1 is incident on a boundary at an angle of 30o. The extent to which the light is bent due to refraction depends, in part, on the index of refraction n2of material 2. Figure 35-32b gives the angle of refraction θ2versus n2for a range of possible n2values, from na=1.30to nb=1.90. What is the speed of light in material 1?

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