Question

In physics lab, you are studying the properties of four transparent liquids. You shine a ray of light (in air) onto the surface of each liquid—A, B, C, and D—one at a time, at a 60.0_ angle of incidence; you then measure the angle of refraction. The table gives your data: Liquid A B C D Ua1 _2 36.4 40.5 32.1 35.2 The wavelength of the light when it is traveling in air is 589 nm.

(a) Find the refractive index of each liquid at this wavelength. Use Table 33.1 to identify each liquid, assuming that all four are listed in the table.

(b) For each liquid, what is the dielectric constant K at the frequency of the 589-nm light? For each liquid, the relative permeability (Km) is very close to unity. (c) What is the frequency of the light in air and in each liquid?

Short answer

a) A is carbon tetrachloride, B is water, C is carbon disulfide and D is Benzene.

b) $K_A$=2.13, $K_B$ =1.77, $K_C$ =2.66, $K_D$=2.25.

c)$f_A=f_B=f_C=f_D=5.1\times {10}^{14}Hz$

Step-by-step solution

Solving part (a) of the problem.

The refractive index of an optical material is expressed by n and represents the speed of light in the vacuum divided by the speed of light in the material. Snell's law is given by an equation in the form

$n_{air}\sin {\theta }_a=n\sin {\theta }_b$

Where $n_{air}$= 1 and n is the refractive index of the liquid. Now, we solve equation (1) for n and use the values for${\theta }_a=60°$ and${\theta }_b$ , for each liquid as next

liquid A: (${\theta }_b=36.4°$)

$n=\frac{\sin {\theta }_a}{\sin {\theta }_b}=\frac{\sin 60°}{\sin 36.4°}=1.46$

we conclude that the liquid is Carbon tetrachloride

liquid B:${\theta }_b=40.5°$

$n=\frac{\sin {\theta }_a}{\sin {\theta }_b}=\frac{\sin 60°}{\sin 40.5°}=1.33$

we conclude that the liquid is Water

liquid C:${\theta }_b=32.1°$

$n=\frac{\sin {\theta }_a}{\sin {\theta }_b}=\frac{\sin 60°}{\sin 32.1°}=1.63$

, we conclude that the liquid is Carbon disulfide

liquid D:${\theta }_b=35.2°$

$n=\frac{\sin {\theta }_a}{\sin {\theta }_b}=\frac{\sin 60°}{\sin 35.2°}=1.50$

we conclude that the liquid is Benzene|

Solving part (b) of the problem.

The dielectric is related to the refractive index by

$K=n^2$

So, for each liquid, we can get the dielectric by

Liquid A

$K_A=n_A^2={\left(1.46\right)}^2=2.13$

Liquid B

$K_B=n_B^2={\left(1.33\right)}^2=1.77$

Liquid C

$K_C={n_c}^2={\left(1.63\right)}^2=2.66$

Liquid D

$K_D=n_D^2={\left(1.50\right)}^2=2.25$

Solving part (c) of the problem.

The frequency of the light doesn't change as the medium changes even the speed of the light change. So, the frequency for all liquids is the same

$f_A=f_B=f_C=f_D$

The frequency is related to the wavelength by

$f=\frac{c}{\lambda }$

Now, plug the values for c and$\lambda$ to get f

$f=\frac{c}{\lambda }=\frac{3\times {10}^{8}m/s}{589\times {10}^{-9}m}=5.1\times {10}^{14}Hz$

Therefore, the frequency for each liquid is

$f_A=f_B=f_C=f_D=5.1\times {10}^{14}Hz$