Chapter 27

$1/1\text{.}\text{333}=0\text{.}\text{750}$

1.49, Polystyrene

0.877 glass to water

$0\text{.}\text{516º}$

$1\text{.}\text{22}\times {\text{10}}^{-6}\text{m}$

600 nm

$2\text{.}\text{06º}$

1200 nm (not visible)

(a) 760 nm

(b) 1520 nm

For small angles $\text{sin}\theta -\text{tan}\theta \approx \theta (\text{in radians})$.

For two adjacent fringes we have,

and

Subtracting these equations gives

450 nm

$5\text{.}\text{97º}$

$8\text{.}\text{99}\times {\text{10}}^3$

707 nm

(a) $\begin{array}{c}\mathrm{ 11.8º},\mathrm{12.5º},\mathrm{14.1º},\mathrm{19.2º}\end{array}$

(b) $\begin{array}{c}\mathrm{ 24.2º},\mathrm{25.7º},\mathrm{29.1º},\mathrm{41.0º}\end{array}$

(c) Decreasing the number of lines per centimeter by a factor of x means that the angle for the x­‐order maximum is the same as the original angle for the first-order maximum.

589.1 nm and 589.6 nm

$\text{28.7º}$

$\text{43}\text{.}\mathrm{2º}$

$\text{90}\text{.}\mathrm{0º}$

(a) The longest wavelength is 333.3 nm, which is not visible.

(b) 333 nm (UV)

(c) $6\text{.}\text{58}\times {\text{10}}^3\text{cm}$

$1\text{.}\text{13}\times {\text{10}}^{-2}\text{m}$

(a) 42.3 nm

(b) Not a visible wavelength

The number of slits in this diffraction grating is too large. Etching in integrated circuits can be done to a resolution of 50 nm, so slit separations of 400 nm are at the limit of what we can do today. This line spacing is too small to produce diffraction of light.

(a) $\text{33}\text{.}\mathrm{4º}$

(b) No

(a) $1\text{.}\text{35}\times {\text{10}}^{-6}\text{m}$

(b) $\text{69}\text{.}\mathrm{9º}$

750 nm

(a) $9\text{.}\text{04º}$

(b) 12

(a) $0\text{.}\text{0150º}$

(b) 0.262 mm

(c) This distance is not easily measured by human eye, but under a microscope or magnifying glass it is quite easily measurable.

(a) $\text{30}\text{.}\mathrm{1º}$

(b) $\text{48}\text{.}\mathrm{7º}$

(c) No

(d) ${\mathrm{2\theta }}_1=(2)(\text{14.5º})=\text{29º},{\theta }_2-{\theta }_1=\text{30}\text{.}\text{05º}-\text{14}\text{.}\mathrm{5º}\text{=}\text{15}\text{.}\text{56º}$. Thus, $\text{29º}\approx (2)(\text{15}\text{.}\text{56º})=\text{31}\text{.}\mathrm{1º}$.

$\text{23}\text{.}\mathrm{6º}$ and $\text{53}\text{.}\mathrm{1º}$

(a) $1\text{.}\text{63}\times {\text{10}}^{-4}\text{rad}$

(b) 326 ly

$1\text{.}\text{46}\times {\text{10}}^{-5}\text{rad}$

(a) $3\text{.}\text{04}\times {\text{10}}^{-7}\text{rad}$

(b) Diameter of $\mathrm{235\; m}$

5.15 cm

(a) Yes. Should easily be able to discern.

(b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations.

532 nm (green)

83.9 nm

620 nm (orange)

380 nm

33.9 nm

$4\text{.}\text{42}\times {\text{10}}^{-5}\text{m}$

The oil film will appear black, since the reflected light is not in the visible part of the spectrum.

$\text{45}\text{.}\mathrm{0º}$

$\text{45}\text{.}7{\text{mW/m}}^2$

$\text{90}\text{.}\mathrm{0\%}$

$I_0$

$\text{48}\text{.}\mathrm{8º}$

$\text{41}\text{.}\mathrm{2º}$

(a) 1.92, not diamond (Zircon)

(b) $\text{55}\text{.}\mathrm{2º}$

$B_2=0\text{.}\text{707}{B}_1$

(a) $2.07\times {10}^{-2}$ °C/s

(b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight.

(a) $\sin \theta =\frac{4.00\times {10}^{-7}}{{10}^{-6}}=0.400\theta =0.400=23.6°$

(b) $\sin \theta =\frac{(2)(4.00\times {10}^{-7})}{{10}^{-6}}0.800\theta =0.800=53.1°$

(c) $\sin 23.9°=\frac{x}{{10}^{-6}\text{ cm}}x=\sin 23.9°({10}^{-6})=4.05\times {10}^{-7}=405\text{ nm}$

(d) $\sin \theta =\frac{4.00\times {10}^{-7}}{2.00\times {10}^{-6}}=0.200\theta =0.200=11.5°$

(e) No.

(b)

(b) and (c)

(b)

(b)

(b)

(d)

(b)

(d)

(b)