Figure 35-56a show two light rays that are initially in phase as they travel upward through a block of plastic, with wavelength 400 nm as measured in air. Light ray ${\mathbf{r}}_{\mathbf{1}}$exits directly into air. However, before light ray ${\mathbf{r}}_{\mathbf{2}}$exits into air, it travels through a liquid in a hollow cylinder within the plastic. Initially the height ${\mathbf{L}}_{\mathbf{liq}}$of the liquid is 40.0 $\mathrm{\mu m}$, but then the liquid begins to evaporate. Let $\mathrm{\theta }$be the phase difference between rays${\mathbf{r}}_{\mathbf{1}}$and ${\mathbf{r}}_{\mathbf{2}}$once they both exit into the air. Figure 35-56b, shows$\mathbf{\theta }$ versus the liquid’s height${\mathbf{L}}_{\mathbf{liq}}$ until the liquid disappears, with $\mathbf{\theta }$given in terms of wavelength and the horizontal scale set by${\mathbf{L}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{40}\mathbf{.}\mathbf{00}\mathbf{\mu m}$.What are (a) the index of refraction of the plastic and (b) the index of refraction of the liquid?

Step by step solution

01

Given in the question.

The height of liquid is$L_{liq}=40.0\mu m$

The height of solid is$L_s=40.0\mu m$

02

 Step 2: Formula of refractive index.

Use the formula for refractive index,

$\mathbf{n}\mathbf{=}\frac{\mathbf{c}}{\mathbf{v}}$

Here, is speed in air,$\mathbf{v}$is speed in medium

03

(a) The index of refraction of the plastic.

According to the question, $L_{iy}=0,\varphi =60\lambda ,L=40.0\mu m$

Thus, write as follows:

$$\begin{gathered} L{n}_{plastic}-L{n}_{air}=60.0\lambda \\ {n}_{plastic}=\frac{60\lambda +L{n}_{air}}{L} \end{gathered}$$

Substitute for $\lambda =400 \text{nm}=400\times {10}^{-9} \text{m,}{n}_{air}=1,L=40\mu m=40\times {10}^{-6} \text{m}$

$$\begin{gathered} n_{plastic}=\frac{60\left(400\times {10}^{-9} \text{m}\right)+\left(40\times {10}^{-6} \text{m}\right)\left(1\right)}{40\times {10}^{-6} \text{m}} \\ =1.6 \end{gathered}$$

Hence, the index of refraction of the plastic is $1.6$.

04

(b) The index of refraction of the liquid.  

Solve as follows:

$$\begin{gathered} L{n}_{liq}-L{n}_{air}=20\lambda \\ {n}_{liq}=\frac{20\lambda +L{n}_{air}}{L} \\ =\frac{20\times \left(400\times {10}^{-9} \text{m}\right)+\left(40\times {10}^{-6} \text{m}\right)\left(1\right)}{40\times {10}^{-6} \text{m}} \\ =1.2 \end{gathered}$$

Hence, the index of refraction of the liquid is $1.2$.

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