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Can a sound wave in air be polarized? Explain.
No light passes through two perfect polarizing filters with perpendicular axes. However, if a third polarizing filter is placed between the original two, some light can pass. Why is this? Under what circumstances does most of the light pass?
Explain what happens to the energy carried by light that it is dimmed by passing it through two crossed polarizing filters.
When particles scattering light are much smaller than its wavelength, the amount of scattering is proportional to $1/{\lambda}^{4}$ . Does this mean there is more scattering for small $\lambda $ than large $\lambda $ ? How does this relate to the fact that the sky is blue?
Using the information given in the preceding question, explain why sunsets are red.
When light is reflected at Brewster’s angle from a smooth surface, it is $\text{100\%}\text{}$ polarized parallel to the surface. Part of the light will be refracted into the surface. Describe how you would do an experiment to determine the polarization of the refracted light. What direction would you expect the polarization to have and would you expect it to be $\text{100\%}\text{}$ ?
What angle is needed between the direction of polarized light and the axis of a polarizing filter to cut its intensity in half?
$\text{45}\text{.}\mathrm{0\xba}$
The angle between the axes of two polarizing filters is $\text{45}\text{.}\mathrm{0\xba}$ . By how much does the second filter reduce the intensity of the light coming through the first?
If you have completely polarized light of intensity $\text{150 W}/{\text{m}}^{2}$ , what will its intensity be after passing through a polarizing filter with its axis at an $\text{89}\text{.}\mathrm{0\xba}$ angle to the light’s polarization direction?
$\text{45}\text{.}7\phantom{\rule{0.25em}{0ex}}{\text{mW/m}}^{2}$
What angle would the axis of a polarizing filter need to make with the direction of polarized light of intensity $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{kW/m}}^{2}$ to reduce the intensity to $10\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ ?
At the end of [link] , it was stated that the intensity of polarized light is reduced to $\text{90}\text{.}\mathrm{0\%}\text{}$ of its original value by passing through a polarizing filter with its axis at an angle of $\text{18}\text{.}\mathrm{4\xba}$ to the direction of polarization. Verify this statement.
$\text{90}\text{.}\mathrm{0\%}\text{}$
Show that if you have three polarizing filters, with the second at an angle of $\text{45\xba}$ to the first and the third at an angle of $\text{90}\text{.}\mathrm{0\xba}$ to the first, the intensity of light passed by the first will be reduced to $\text{25}\text{.}\mathrm{0\%}\text{}$ of its value. (This is in contrast to having only the first and third, which reduces the intensity to zero, so that placing the second between them increases the intensity of the transmitted light.)
Prove that, if $I$ is the intensity of light transmitted by two polarizing filters with axes at an angle $\theta $ and $I\prime $ is the intensity when the axes are at an angle $\text{90.0\xba}-\mathrm{\theta ,}$ then $I+I\prime ={I}_{\mathrm{0,}}$ the original intensity. (Hint: Use the trigonometric identities $\text{cos}\phantom{\rule{0.25em}{0ex}}(\mathrm{90.0\xba}-\theta )=\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ and ${\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\theta +{\text{sin}}^{2}\phantom{\rule{0.25em}{0ex}}\theta =1.$ )
${I}_{0}$
At what angle will light reflected from diamond be completely polarized?
What is Brewster’s angle for light traveling in water that is reflected from crown glass?
$\text{48}\text{.}\mathrm{8\xba}$
A scuba diver sees light reflected from the water’s surface. At what angle will this light be completely polarized?
At what angle is light inside crown glass completely polarized when reflected from water, as in a fish tank?
$\text{41}\text{.}\mathrm{2\xba}$
Light reflected at $\text{55}\text{.}\mathrm{6\xba}$ from a window is completely polarized. What is the window’s index of refraction and the likely substance of which it is made?
(a) Light reflected at $\text{62}\text{.}\mathrm{5\xba}$ from a gemstone in a ring is completely polarized. Can the gem be a diamond? (b) At what angle would the light be completely polarized if the gem was in water?
(a) 1.92, not diamond (Zircon)
(b) $\text{55}\text{.}\mathrm{2\xba}$
If ${\theta}_{\text{b}}$ is Brewster’s angle for light reflected from the top of an interface between two substances, and ${\theta \prime}_{\text{b}}$ is Brewster’s angle for light reflected from below, prove that ${\theta}_{\text{b}}+{\theta \prime}_{\text{b}}=\text{90}\text{.}\mathrm{0\xba.}$
Integrated Concepts
If a polarizing filter reduces the intensity of polarized light to $\text{50}\text{.}\mathrm{0\%}\text{}$ of its original value, by how much are the electric and magnetic fields reduced?
${B}_{2}=0\text{.}\text{707}\phantom{\rule{0.25em}{0ex}}{B}_{1}$
Integrated Concepts
Suppose you put on two pairs of Polaroid sunglasses with their axes at an angle of $\text{15}\text{.}\mathrm{0\xba}$ . How much longer will it take the light to deposit a given amount of energy in your eye compared with a single pair of sunglasses? Assume the lenses are clear except for their polarizing characteristics.
Integrated Concepts
(a) On a day when the intensity of sunlight is $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{kW}/{\text{m}}^{2}$ , a circular lens 0.200 m in diameter focuses light onto water in a black beaker. Two polarizing sheets of plastic are placed in front of the lens with their axes at an angle of $\text{20}\text{.}\mathrm{0\xba.}$ Assuming the sunlight is unpolarized and the polarizers are $\text{100\%}\text{}$ efficient, what is the initial rate of heating of the water in $\text{\xbaC}/\text{s}$ , assuming it is $\text{80}\text{.}\mathrm{0\%}\text{}$ absorbed? The aluminum beaker has a mass of 30.0 grams and contains 250 grams of water. (b) Do the polarizing filters get hot? Explain.
(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.
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