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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 r2 interfere, 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,n2, and n3, the type of interference, the thin-layer thickness L in nanometers, and the wavelength λin nanometers of the light as measured in air. Where is 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 wavelength of light is 528 nm.

Step by step solution

01

Interference in thin films:

The bright colors reflected from thin oil on water and soap bubbles are the result of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colors can be seen.

For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

2L=m+12λn2 â¶Ä‰â¶Ä‰m=0, 1, 2,...(Maxima—bright film in the air)

2L=mλn2 â¶Ä‰â¶Ä‰m=1, 2, 3,..(Minima)

Where, λ is the wavelength of the light in air, L is its thickness, and n2 is the film’s refraction index.

02

Determine the wavelength of light:

Here, in this case, light travels in a medium with n1=1.68 and incident on the thin layer whose refractive index is n2=1.59 and the reflected light has no phase change as the light is reflected off the rarer medium. And then, the refracted light gets reflected of the third surface n3=1.50 while traveling through the film. This, too, does not result in phase change. The phase difference between r1 and r2 is zero. As a result, the condition for destructive interference or minimum intensity is

2L=m+12λn2 (destructive)

The wavelength of the reflected light is as below.

For m=0:

λ=4Ln22m+1=4(415 nm)(1.59)1=2639 nm

For m=1:

λ=4Ln22m+1=2639 nm3=880 nm

For m=2:

λ=4Ln22m+1=2639 nm5=528 nm

As 528 nmis in the visible range, the wavelength of reflected light is 528 nm.

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Most popular questions from this chapter

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 and interfere, r3and r4here 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 refraction n1,n2and n3the type of interference, the thin-layer thickness Lin nanometers, and the wavelength in nanometers 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. 35-37, two isotropic point sources S1 and S2 emit identical light waves in phase at wavelengthλ. The sources lie at separation on an x axis, and a light detector is moved in a circle of large radius around the midpoint between them. It detects 30points of zero intensity, including two on the xaxis, one of them to the left of the sources and the other to the right of the sources. What is the value of dλ?

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 the double-slit experiment of Fig. 35-10, the viewing screen is at distance D=4.00m, point P lies at distance role="math" localid="1663143982922" y=20.5cmfrom the center of the pattern, the slit separation d is 4.50mm, and the wavelength λis 580 nm. (a) Determine where point P is in the interference pattern by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies. (b) What is the ratio of the intensitylPat point P to the intensitylcen at the centerof the pattern?

Suppose that Young’s experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits. How far apart are the bright fringes near the center of the interference pattern?

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