## Chapter 3

### Check Your Understanding

$3.63\text{\xb0}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}7.27\text{\xb0}$, respectively

a. too small; b. up to $8\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}$

### Conceptual Questions

Because both the sodium lamps are not coherent pairs of light sources. Two lasers operating independently are also not coherent so no interference pattern results.

Monochromatic sources produce fringes at angles according to $d\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =m\lambda $. With white light, each constituent wavelength will produce fringes at its own set of angles, blending into the fringes of adjacent wavelengths. This results in rainbow patterns.

Differing path lengths result in different phases at destination resulting in constructive or destructive interference accordingly. Reflection can cause a $180\text{\xb0}$ phase change, which also affects how waves interfere. Refraction into another medium changes the wavelength inside that medium such that a wave can emerge from the medium with a different phase compared to another wave that travelled the same distance in a different medium.

The surface of the ham being moist means there is a thin layer of fluid, resulting in thin-film interference. Because the exact thickness of the film varies across the piece of ham, which is illuminated by white light, different wavelengths produce bright fringes at different locations, resulting in rainbow colors.

Other wavelengths will not generally satisfy $t=\frac{\lambda \text{/}n}{4}$ for the same value of *t* so reflections will result in completely destructive interference. For an incidence angle $\theta $, the path length inside the coating will be increased by a factor $1\text{/}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta $ so the new condition for destructive interference becomes $\frac{t}{\text{cos}\phantom{\rule{0.2em}{0ex}}\theta}=\frac{\lambda \text{/}n}{4}$.

### Problems

$5.77\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-7}}\phantom{\rule{0.2em}{0ex}}\text{m}=577\phantom{\rule{0.2em}{0ex}}\text{nm}$

a. $0.40\text{\xb0},0.53\text{\xb0};$ b. $4.6\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-3}}\phantom{\rule{0.2em}{0ex}}\text{m}$

$8.39\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-8}}\phantom{\rule{0.2em}{0ex}}\text{m}=83.9\phantom{\rule{0.2em}{0ex}}\text{nm}$

a. Assuming *n* for the plane is greater than 1.20, then there are two phase changes: 0.833 cm. b. It is too thick, and the plane would be too heavy. c. It is unreasonable to think the layer of material could be any thickness when used on a real aircraft.

$4.55\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{m}$

$D=2.53\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}\phantom{\rule{0.2em}{0ex}}\text{m}$

### Additional Problems

$0.29\text{\xb0}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}0.86\text{\xb0}$

$52.7\phantom{\rule{0.2em}{0ex}}\mu \text{m}$ and $53.0\phantom{\rule{0.2em}{0ex}}\mu \text{m}$

### Challenge Problems

The path length must be less than one-fourth of the shortest visible wavelength in oil. The thickness of the oil is half the path length, so it must be less than one-eighth of the shortest visible wavelength in oil. If we take 380 nm to be the shortest visible wavelength in air, 33.9 nm.

$4.42\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}\phantom{\rule{0.2em}{0ex}}\text{m}$

for one phase change: 950 nm (infrared); for three phase changes: 317 nm (ultraviolet); Therefore, the oil film will appear black, since the reflected light is not in the visible part of the spectrum.