Q65P Transmission through thin layers... [FREE SOLUTION]

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Chapter 35: Q65P (page 1077)

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 $r_{3}$ (the light does not reflect inside material 2) and $r_{4}$ (the light reflects twice inside material 2). The waves of $r_{3}$ and $r_{4}$ 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 $n_{1}, n_{2}$ and $n_{3}$ the type of interference, the thin-layer thickness $L$ in nanometers, and the wavelength $位$ in nanometers of the light as measured in air. Where $位$ is 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.

Short Answer

Expert verified

The thickness of the thin layer is $339 nm$ .

Step by step solution

Given Data:

The refractive index of first medium is $n_{1} = 1.60$ .
The refractive index of the thin film is $n_{2} = 1.40$
The refractive index of the third medium is $n_{3} = 1.80$ .
The minimum intensity occurs at $位_{\min} = 632 nm$ .

Interference of light through thin films:

Light that is incident normally on thin films is reflected from both the front and back surfaces, causing interference of the reflected light. When constructive interference happens, it produces bright reflected light, and when entirely destructive interference occurs, it produces a dark region.

Define the wavelength:

The interference of the transmitted rays is similar to the interference of the reflection of light. Here in this case, as $n_{1} > n_{2}$ and $n_{2} > n_{3}$ the two transmitted rays have zero phase angle difference because the ray $r_{4}$ will undergo $180^{鈭�}$ phase change twice.