In Fig 35-59, an oil drop (n=1.20) floats on the surface of water (n=1.33) and is viewed from overhead when illuminated by sunlight shinning vertically downward and reflected vertically upward. (a) Are the outer (thinnest) regions of the drop bright or dark? The oil film displays several spectra of colors. (b) Move from the rim inward to the third blue band and using a wavelength of 475 nm for blue light, determine the film thickness there. (c) If the oil thickness increases, why do the colors gradually fade and then disappear?

Short Answer

Expert verified

(a) The film thickness is t=3λ2μ0.

(b) The oil thickness is594nm .

(c) The fundamental cause of color fading and eventual disappearance when oil thickness rises is that the colored bands start to blend too much to be easily differentiated. Furthermore, there would be too much space between the two reflecting surfaces for the light that reflected from them to be coherent.

Step by step solution

01

Given in the question.

The formula for the reflective index of oil:

μ0=1.20

Reflection index of water,μw=1.33

Here, λis wavelength of light in air/vacuum.

02

Definition of refractive index.

The reflective index is the factor of speed and wavelength of the radiation with respect to vacuum values:

λ0=λμ0

03

 Step 3: (a) Outer region of the drop.

Phase difference as follows:

Δ=2πλ0Δx=2πλ0·2t

Here,t -thickness of oil.

For constructive reflection as follows;

Δ=2πλ0·2t=2nπ=2μt=nλ

Here, 2t-potential difference between incident and reflected,n=0,1,2,3,...

For destructive reflection as follows:

Δ=2μt2πλ0=2n+1π2μt=n+12λ

Near rim of drop,t<λ4μ ie, constructive outer region –bright.

Hence, the film thickness ist=3λ2μ0

04

(b) Calculate the film thickness. 

The third band from rim, 2μt=3λfilm thickness

t=3λ2μ0=3×475nm2×1.20=594nm

As the oil thickness increases, colours gradually fade and disappear.

Hence, the oil thickness is 594nm.

05

(c) The increase in the thickness. 

The fundamental cause of color fading and eventual disappearance when oil thickness rises is that the colored bands start to blend too much to be easily differentiated. Furthermore, there would be too much space between the two reflecting surfaces for the light that reflected from them to be coherent.

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Most popular questions from this chapter

Figure 35-29 shows the transmission of light through a thin film in the air by a perpendicular beam (tilted in the figure for clarity). (a) Did rayr3undergo a phase shift due to reflection? (b) In wavelengths, what is the reflection phase shift for rayr4? (c) If the film thickness is L, what is the path length difference between raysr3andr4?

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 r3and r4interfere, 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 refraction n1,n2and n3the type of interference, the thin-layer thickness Lin nanometers, and the wavelength λin nanometers of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where L is missing, give the second least thickness or the third least thickness as indicated.

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 λin 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.

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.

Two waves of light in air, of wavelength λ=600.0nm, are initially in phase. They then both travel through a layer of plastic as shown in Fig. 35-36, with L1=4.00μm, L2=3.50μm, n1=1.40, n2=1.60and. (a) What multiple of λgives their phase difference after they both have emerged from the layers? (b) If the waves later arrive at some common point with the same amplitude, is their interference fully constructive, fully destructive, intermediate but closer to fully constructive,or intermediate but closer to fully destructive?

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