Recall that for all electromagnetic waves, including light waves, the direction of the \(\texttip{\vec{E}}{E_vec}\) field is thedirec

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Recall that for all electromagnetic waves, including light waves, the direction of the \(\texttip{\vec{E}}{E_vec}\) field is thedirection of polarization and is perpendicular to the propagation direction. When working with polarizers, you are reallydealing with components of \(\texttip{\vec{E}}{E_vec}\) parallel and perpendicular to the polarizing axis. Everything you knowabout components of vectors is applicable here.SET UP the problem using the following steps:1. Just as for problems in geometric optics, you should always start by drawing a large, neat diagram. Label allknown angles, including the angles of any and all polarizing axes.2. Determine the target variables.EXECUTE the solution as follows:1. Keep in mind that a polarizer lets pass only electric-field components parallel to its polarizing axis.2. If the incident light is linearly polarized and has amplitude \(\texttip{E}{E}\) and intensity \(\texttip{I_{\rm max}}{I_max}\), the light that passes through an ideal polarizer has amplitude \(E \cos{\phi}\) and intensity \(I_{\rmmax} \cos ^2{\phi}\), where \(\texttip{\phi }{phi}\) is the angle between the incident polarization direction and thefilter’s polarizing axis.3. Unpolarized light is a random mixture of all possible polarization states. So on the average, it has equalcomponents in any two perpendicular directions. When passed through an ideal polarizer, unpolarized lightbecomes linearly polarized light with half the incident intensity. Partially linearly polarized light is a superpositionof linearly polarized and unpolarized light.4. The intensity (average power per unit area) of a wave is proportional to the squareof its amplitude. If you findthat two waves differ in amplitude by a certain factor, their intensities differ by the square of that factor.EVALUATE your answer:Check your answer for any obvious errors. If your results say that light emerging from a polarizer has greater intensity thanthe incident light, something’s wrong: A polarizer can’t add energy to a light wave.IDENTIFY the relevant conceptsThis problem involves a set of two polarizing filters: the first polarizer, on which unpolarized light shines, and the secondpolarizer, on which linearly polarized light shines. Recall that for all electromagnetic waves, including light waves, thedirection of the \(\texttip{\vec{E}}{E_vec}\) field is the direction of polarization.

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6 months 2021-07-29T20:43:59+00:00 1 Answers 2 views 0

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    2021-07-29T20:45:20+00:00

    Answer:

    Polarization is the attribute that a wave’s oscillations have a definite direction relative to the direction of propagation of the wave. (This is not the same type of polarization as that discussed for the separation of charges.) Waves having such a direction are said to be polarized. For an EM wave, we define the direction of polarization to be the direction parallel to the electric field. Thus we can think of the electric field arrows as showing the direction of polarization, as in Figure 2.

    To examine this further, consider the transverse waves in the ropes shown in Figure 3. The oscillations in one rope are in a vertical plane and are said to be vertically polarized. Those in the other rope are in a horizontal plane and are horizontally polarized. If a vertical slit is placed on the first rope, the waves pass through. However, a vertical slit blocks the horizontally polarized waves. For EM waves, the direction of the electric field is analogous to the disturbances on the ropes.

    Explanation:

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