Answer to Question #108947 in Electricity and Magnetism for Larry

Question

An incident plane electromagnetic wave with frequency ω, wavenumber kI = kIz-hat, and (real) amplitude E0I travels through a vacuum and strikes a dielectric material (with index of refraction n) at normal incidence. The amplitude of the reflected wave is measured to be E0R = (1/2)E0I

Find the index of refraction n.
Find the wavelength of the transmitted wave, λT
Explain (in detail) why the incident and reflected waves have the same wavelength, λI = λR. Write down any equations that support your response.

Answer 1

As per the given question,

frequency =ω

wave number $=k$

amplitude $=E_o$

Let the speed of light is c,

i) Amplitude of the reflected wave $\dfrac{E_o}{2}$

Speed of the wave (v)=$E_o \omega$

$n=\dfrac{c}{v}=\dfrac{c}{E_o\omega}$

ii) We know that, wave number ,$k=\dfrac{2\pi}{\lambda_T}$

$\lambda_T=\dfrac{2\pi}{k}$

In the reflection, the angle of incident and angle of reflection always be same.

So, the wave number must be same,

so, angle of reflection is equal to the angle of incidence θi, and the magnitude of the incident and reflected wave vectors are equal (i.e., the incident and reflected waves have the same wavelength λi = λr),

So, $\dfrac{2\pi}{\lambda_i}=k_i =k_r =\dfrac{2\pi}{\lambda_r}$

Hence, $\theta _i=\theta_r$