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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 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, n2, and 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 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 wavelength of the reflected light is 455nm.

Step by step solution

01

Interference in thin films:

Light interference is the cause of the vivid colors that thin oil on water and soap bubbles reflect. These vivid hues can be seen because of the light reflection from the thin film's front and rear surfaces' constructive interference.

For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:(Maxima—bright film in the air)

2L=m+12λn2m=0,1,2,...(Maxima—bright film in the air)

2L=mλn2m=1,2,3,..(Minima)

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

02

Determine the wavelength of light:

Here, in this case, light travels in a medium with n1=1.32and incident on the thin layer whose refractive index is n2=1.75and the reflected light has phase change as the light is reflected off the denser medium. And then, the refracted light gets reflected of the back surface n3=1.39while traveling through the film.

This results in no phase change. The total phase difference between r1and r2 is 180o.

As a result, the condition for constructive interference or maximum intensity is,

2L=m+12λn2λ=4Ln22m+1

The wavelength for first few orders is as below.

For : m=0

localid="1663011607370" λ=4325nm1.7520+1=2275nm

For:m=1

localid="1663011956572" λ=4(325mm)(1.75)2(1)+1=758mm

For :m=2

λ=4(325mm)(1.75)2(2)+1=455nm

As 455nmis in the visible range, the wavelength of the light is 455nm.

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Most popular questions from this chapter

In Fig. 35-34, a light ray is an incident at angle θ1=50°on a series of five transparent layers with parallel boundaries. For layers 1 and 3 , L1=20μm , L2=25μm, n1=1.6and n3=1.45. (a) At what angle does the light emerge back into air at the right? (b) How much time does the light take to travel through layer 3?

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