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White light is sent downward onto a horizontal thin film that is sandwiched between two materials. The indexes of refraction are 1.80for the top material, 1.70for the thin film, and 1.50for the bottom material. The film thickness is5×10-7m . Of the visible wavelengths (400 to 700nm ) that result in fully constructive interference at an observer above the film, which is the (a) longer and (b) shorter wavelength? The materials and film are then heated so that the film thickness increases. (c) Does the light resulting in fully constructive interference shift toward longer or shorter wavelengths?

Short Answer

Expert verified

(a) The longer wavelength for constructive interference is 567nm.

(b) The shorter wavelength for constructive interference is 425nm.

(c) The resulting light in constructive interference shifts towards longer wavelength.

Step by step solution

01

Identification of given data

The index of refraction for top material is nt=1.80

The index of refraction for bottom material is nb=1.50.

The index of refraction for thin film is n=1.70.

The thickness of thin film is t=5×10-7m.

02

Understanding the concept

The thickness of the thin film in interference varies with the index of refraction, wavelength of light and order of fringe.

03

(a) Determination of longer wavelengths in constructive interference for visible range

The wavelength of the light for constructive interference is given as:

λm=2ntm (i)

Substitute m=1in the above equation (i).

λ1=21.705×10-7m1=17×10-7m=17×10-7m1nm10-9m=1700nm

Substitute m=2in the above equation (ii).

λ2=21.705×10-7m2=8.50×10-7m=8.50×10-7m1nm10-9m=850nm

Substitute m=3in the above equation (i).

λ3=21.705×10-7m3=5.67×10-7m=5.67×10-7m1nm10-9m=567nm

The wavelengths for first and second order fringes are more than the maximum value of wavelength of visible range of light. The wavelength for third order fringe is longer wavelength for constructive interference in visible range because it is next longer wavelength inside visible range of light.

Therefore, the longer wavelength for constructive interference is 567nm.

04

(b) Determination of longer wavelengths in constructive interference for visible range

Substitute m=4in the above equation (i).

λ4=21.705×10-7m4=4.25×10-7m=4.25×10-7m1nm10-9m=425nm

Substitute m=5in the above equation (i).

λ5=21.705×10-7m5=3.40×10-7m=3.40×10-7m1nm10-9m=340nm

The wavelength for fifth order fringe is less than the minimum value of wavelength of visible range of light. The wavelength for fourth order fringe is shorter wavelength for constructive interference in visible range because it is next shorter wavelength inside visible range of light.

Therefore, the shorter wavelength for constructive interference is 425nm.

05

(c) Determination of shift of resulting light in constructive interference

The index of refraction of thin film remains unaffected by the variation in temperature so the resulting light in constructive interference shifts towards longer wavelength.

Therefore, the resulting light in constructive interference shifts towards longer wavelength.

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Most popular questions from this chapter

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?

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.

A disabled tanker leaks kerosene n=1.20into the Persian Gulf, creating a large slick on top of the watern=1.30). (a) If you are looking straight down from an airplane, while the Sun is overhead, at a region of the slick where its thickness is460nm, for which wavelength(s) of visible light is the reflection brightest because of constructive interference? (b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest?

Transmission through thin layers. In Fig. 35-43, 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.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of r3andr4interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refractionn1,n2andn3the type of interference, the thin-layer thicknessLin nanometers, and the wavelengthλin nanometers of the light as measured in air. Whereλis missing, give the wavelength that is in the visible range. WhereLis missing, give the second least thickness or the third least thickness as indicated.

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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