Home Page Textbooks Physics: Principles and Problems Page 447: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 447: Section Review

Exercise 18

Step 1

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You get white light when you combine the three primaries

We already have blue, we need red and green.

The mixture of red and green is yellow, therefore we should add yellow to blue

Exercise 19

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The magenta and yellow pigments can be used to make red. The yellow pigment subtracts blue and magenta pigment subtracts green. Both of them don’t subtract red; hence, the mixture of the two will show a red color.

Exercise 20

Solution 1

Solution 2

Step 1

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a) Yellow

b) Yellow

c)Black

Result

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a) $textit$$text{color{yellow} Yellow $ $ \\ b) $textit $color{yellow} Yellow $ $ \\ c) $textit $color{default} Black $ $}$

Step 1

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a) Any object that’s illuminated by the white light will reflect the same color to our eye, which means the yellow banana will appear **yellow**.
b) If the yellow banana is illuminated by the red and green light it will look **yellow**. Because yellow color reflects both green and red.
c) Yellow doesn’t contain blue color which means the banana will appear **black** on the blue light.

Result

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a) Yellow
b) Yellow
c) Black

Exercise 21

Step 1

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The speed of light in a vacuum is $c$, or $3 cdot 10^{8} hspace{0.5mm} frac{m}{s}$. The speed of light in a medium, as water or air, is $v=frac{c}{n}$ where $n$ is the refractive index. Also, we know that the equation with a frequency and a wavelength in a medium is $f= frac{ lambda_{ text{medium}}}{ v}$ The equation with a frequency and a wavelength in a vacuum is $f= frac{ lambda_{ text{vacuum}}}{ c}$ The frenquencies in both cases are equal, so we can write

$$
begin{align*}
frac{ lambda_{ text{medium}}}{ v}&= frac{ lambda_{ text{vacuum}}}{ c}\
frac{ lambda_{ text{medium}}}{ c/n}&= frac{ lambda_{ text{vacuum}}}{ c}\
lambda_{ text{medium}}&= frac{ lambda_{vacuum}}{n}
end{align*}
$$

The refractive index is greater than $1$, so the wavelength in some medium is greater than the wavelength in a vacuum. So, the wavelength is changed.

Result

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The wavelength is changed.

Exercise 22

Solution 1

Solution 2

Step 1

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See if the glasses reduce glare from the
reflective surfaces, such as windows or
roadways. Polarization of light allows
photographers to photograph objects
while eliminating glare.

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Polarized sunglasses should eliminate the glare. That can be checked by looking at any reflective surface such as water, which produces glare. Then you need to rotate the sunglasses about $60^circ$ to check if the glare is reduced or if it completely disappeared. This is just one of the ways to be sure the sunglasses are polarized.

Exercise 23

Step 1

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The spectral lines of the emissions of
known atoms are blue-shifted in the
light we see coming from Andromeda.
Andromeda would be moving toward us
due to gravitational attraction. This
gravitational attraction could be due to
the mass of the Milky Way or other
objects located near the Milky Way.

Result

2 of 2

$$
textit{color{#c34632} $See$ $Explanation$}
$$

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