A plane wave of monochromatic light is incident normally on a uniform thin film of oil that covers a glass plate. The wavelength of the source can be varied continuously. Fully destructive interference of the reflected light is observed for wavelengths of 500nmand 700nmand for no wavelengths in between. If the index of refraction of the oil is 1.30and that of the glass is 1.50, find the thickness of the oil film.

Short Answer

Expert verified

The thickness of the oil film is 673nm.

Step by step solution

01

Given Data.

  • The refractive index of the oil film isn2=1.30
  • The refractive index of the glass isn3=1.50
  • The minimum intensity is observed at wavelengths role="math" localid="1663025634127" λmin1=500nmandrole="math" localid="1663025642919" λmin2=700nm
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)

whereλis the wavelength of the light in air, Lis its thickness, and n2is the film’s refractive index.

Here, in this case, light travels in an air medium and incident on the oil film whose refractive index is higher than air. And then, the light gets reflected of the thick glass while traveling through oil, and again the refractive index is higher than acetone. As a result, the condition for destructive interference is

2L=m+12λ600n2 (Destructive)

03

Determining thickness of the oil film.

For destructive interference of 500nmreflected light, the value of thickness for the first few orders is

m=0;L=0+12500nm21.30=96nmm=1;L=1+12500nm21.30=289nmm=2;L=2+12500nm21.30=481nmm=3;L=3+12500nm21.30=673nm

For destructive interference of 700nmreflected light, the value of thickness for the first few orders is

role="math" localid="1663026287442" m=0;L=0+12700nm21.30=135nmm=1;L=1+12700nm21.30=404nmm=2;L=2+12700nm21.30=481nmm=3;L=3+12500nm21.30=673nm

The least common thickness for500nm and 700nmreflected light is 673nm. Hence the thickness of the oil film is 673nm.

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Most popular questions from this chapter

A Newton’s rings apparatus is to be used to determine the radius of curvature of a lens . The radii of the nth and (n+20th)bright rings are found to be 0.162cm and 0.368cm, respectively, in light of wavelength 546nm. Calculate the radius of curvature of the lower surface of the lens.

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 r3 and r4 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1, n2, and n3, the type of interference, the thin-layer thickness L in nanometers, and the wavelength λ in nanometers of the light as measured in air. Where λ 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.

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.

A thin film suspended in air is 0.410 μmthick and is illuminated with white light incident perpendicularly on its surface. The index of refraction of the film is 1.50. At what wavelength will visible light that is reflected from the two surfaces of the film undergo fully constructive interference?

Figure 35-25 shows two sources s1 and s2 that emit radio waves of wavelengthλin all directions. The sources are exactly in phase and are separated by a distance equal to 1.5λ . The vertical broken line is the perpendicular bisector of the distance between the sources.

(a) If we start at the indicated start point and travel along path 1, does the interference produce a maximum all along the path, a minimum all along the path, or alternating maxima and minima? Repeat for

(b) path 2 (along an axis through the sources) and

(c) path 3 (along a perpendicular to that axis).

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