As we learn in physical optics, thin-film interference can cause some wavelengths of light to be strongly reflected while others not reflected at all. Neglecting absorption all light has to go one way or the other, so wavelengths not reflected are strongly transmitted. (a) For a film, of thickness t surrounded by air, what wavelengths λ (while they are within the film) will be strongly transmitted? (b) What wavelengths (while they are “over” the barrier) of matter waves satisfies condition (6-14)? (c) Comment on the relationship between (a) and (b).

It is shown in thethin-film interference is a phenomenon when a light beam is reflected from a thin film whose width is comparable to the wavelength of the incident light.

Consider the expression to determine the wavelength as:

$2\mathrm{\mu t}\mathrm{cosr}=\mathrm{n\lambda }$ ….. (1)

Here,data-custom-editor="chemistry" $\mathrm{\mu }$= refractive index

data-custom-editor="chemistry" $\mathrm{\lambda }$= wavelength

r =angle of incidence

n =order of interference

t = thickness of air

Consider the expression for the kinetic energy as:

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ …… (2)

Here,

E = Energy (Kinetic) of photon

h = Plank’s constant

c = Speed of light

Write the expression in terms of the potential energy as:

$\mathrm{E}={\mathrm{U}}_{\mathrm{o}}+\frac{{\mathrm{n}}^2{\mathrm{\pi }}^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}$

$\begin{array}{rcl}\sqrt{\frac{2\mathrm{m}\left(\mathrm{E}-{\mathrm{U}}_{\mathrm{o}}\right)}{h^2}}& =& n\mathrm{\pi }\\ & =& \frac{2\mathrm{\pi }}{{\mathrm{\lambda }}^{\text{'}}}\end{array}$ ….. (3)

Here

U_o = Potential energy

L = Width of the potential barrier

m = mass of the particle

From equation (1), you have,

$\mathrm{\lambda }=\frac{2\mathrm{\mu t}\cos \mathrm{r}}{\mathrm{n}}$

Putting cos r=1 and $\mathrm{\mu }=1$for air:

$\mathrm{\lambda }=\frac{2\mathrm{t}}{\mathrm{n}}$

Hence, wavelength for the strong transmittance is given by$\mathrm{\lambda }=\frac{2\mathrm{t}}{\mathrm{n}}$.

From the condition 6-14, rewrite the equation (3) write as,:

$\mathrm{\lambda }=\frac{2\mathrm{L}}{\mathrm{n}}$

Hence, wavelengths satisfying the condition 6-14 can be given as $\mathrm{\lambda }=\frac{2\mathrm{L}}{\mathrm{n}}$.

From the answers to parts (a) and (b), in both the equations you get to know that wavelength depends on the dimension of the quantity and varies inversely with the order.

Hence, they are essentially identical.