Physics Problems & Solutions: Theory: Optics

Pulses along a rope when reflected off a fixed-end are inverted, and when reflected off a free-end, remain in phase. Light waves also exhibit these behaviors when they encounter an interface between two mediums.

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These reflective properties are critical to our understanding of the colors in such thin films as soap bubbles, coatings on camera lenses, colors in a butterfly's wings or peacock's feathers, or oil spills.

Below is a diagram of a thin film and the light rays associated with the reflections and refractions as light impinges on the film.

When ray 1 strikes the top interface, some of the light is partially reflected, ray 2, and the rest is refracted, ray 3.

When ray 3 strikes the bottom interface, some of it is reflected, ray 4, and the remainder is refracted, ray 6.

When ray 4 strikes the top interface from underneath, some is reflected (not shown) and some is refracted, ray 5.

It is the interference between rays 2 and 5 that produces a thin film's color when the film is viewed from above.

The refracted rays remain in-phase with their initial rays.

A good method for analyzing a thin-film problem involves these steps:

Step 1
Find the shift for the wave reflecting off the top surface of the film, $\Delta_a$

$\\n_1 > n_2 \rightarrow \Delta_a = 0 \\n_1 < n_2 \rightarrow \Delta_a = \lambda/2$

Step 2
Find the shift for the wave reflecting off the film's bottom surface, $\Delta_b$
$\\n_2 > n_3 \rightarrow \Delta_b = 2t \\n_2 < n_3 \rightarrow \Delta_b = 2t+\frac{\lambda}{2}$

Step 3
Calculate the relative shift by subtracting the individual shifts:
$\Delta = \Delta_b-\Delta_a$

Step 4
Set the relative shift equal to the condition for constructive interference, or the condition for destructive interference.

Constructive Interference: $\Delta = m\lambda$
Destructive Interference: $\Delta = \bigg(m+\frac{1}{2}\bigg)\lambda$
with

Step 5
Rearrange the equation.

Step 6
Since we are dealing with the behavior of the light in the thin film, we must ALWAYS use the light's wavelength IN THE FILM, $\lambda_n$ . The wavelength in a medium whose refractive index, n is : $\lambda_n=\frac{\lambda}{n}$

Step 7
Solve

Source: physicslab.org, physics.bu.edu

Example:

A thin soap film is formed by dipping a plastic rectangular wand into a solution of soapy water which has a refractive index of 1.4. When viewed in daylight, one portion of the film reflects blue light of wavelength 475 nm. Estimate the minimum thickness of that section of the film.

Will ray 2 be in-phase or out-of-phase with ray 1?

Air to Soap, $n_{air} < n_{soap} \rightarrow$ fixed-end reflector $\rightarrow$ rays 1 and 2 out-of-phase with a phase inversion of

Will ray 5 be in-phase or out-of-phase with ray 1?

Ray 5: Soap to Air, $n_{soap} > n_{air} \rightarrow$ free-end reflector, in-phase with ray 2, also with 3 and 4.

Since refracted rays remain in-phase with their initial rays, it means ray 1 is also in-phase.

Thus, rays 1 and 5 are in-phase with no phase inversion.

To find minimum thickness, $t_\text{min}$

Step 1.
Air to Soap, $n_a < n_s \rightarrow \Delta_a=\lambda/2$

Step 2.
Soap to Air, $n_s > n_a \rightarrow \Delta_b=2t$ (no phase difference)

Step 3.
$\Delta = \Delta_b - \Delta_a=2t - \frac{\lambda}{2}$

Step 4.
Blue light $\rightarrow$ Constructive Interference, $\Delta=m\lambda$

Step 5.
$m\lambda_s = 2t - \frac{\lambda_s}{2} \rightarrow 2t = \bigg(m+\frac{1}{2}\bigg)\lambda$

Step 6.
$\lambda_s = \frac{\lambda_a}{n_s} = \frac{475}{1.4} = 339 \text{ nm}$

Step 7.
$2t = \left(m+\frac{1}{2}\right)\lambda_s$
with

(minimum thickness)

$t = \frac{1}{2}\bigg(0+\frac{1}{2}\bigg)339 = \frac{339}{4}=84.8 \text{ nm}$

Physics Problems & Solutions

Theory: Optics - Thin Film

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