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Lecture 13 Electromagnetic Waves Chp. 34 Thursday Morning

Lecture 13 Electromagnetic Waves Chp. 34 Thursday Morning. Cartoon -. Opening Demo - Warm-up problem Physlet Topics Light is a Electromagnetic wave eye sensitivity Traveling E/M wave - Induced electric and induced magnetic amplitudes Energy transport Poynting vector

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Lecture 13 Electromagnetic Waves Chp. 34 Thursday Morning

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  1. Lecture 13 Electromagnetic Waves Chp. 34Thursday Morning • Cartoon -. Opening Demo - Warm-up problem • Physlet • Topics • Light is a Electromagnetic wave • eye sensitivity • Traveling E/M wave - Induced electric and induced magnetic amplitudes • Energy transport Poynting vector • Pressure produced by E/M wave • Polarization • Reflection, refraction,Snell’s Law, Internal reflection • Prisms and chromatic dispersion • Polarization by reflection

  2. Electromagnetic spectrum

  3. Eye Sensitivity to Color

  4. Production of Electromagnetic waves

  5. How the fields vary at a Point P in space as the wave goes by

  6. Electromagnetic Wave

  7. Properties of a plane wave of light E and B are always perpendicular to the direction of travel E is perpendicular to B E X B = the direction of travel E and B both vary with the same frequency and in phase. Speed of wave is independent of speed of observer Wave doesn’t need a medium to travel in.

  8. A point source of light is one that emits isotropically and the intensity of it falls off as 1/r2 Let P be the power of the source in joules per sec. Then the intensity of light at a distance r is I = P/4pr2

  9. The Poynting vector S S has units of energy/time per unit area or Watt/m2 E is perp. to B and in the energy flows in the direction of the wave. Since B=E/c, we get for the instantaneous power of the wave Intensity I of the wave is defined as Savg

  10. (c) The power of the source is 17. The maximum electric field at a distance of 10 m from an isotropic point light source is 2.0 V/m. Calculate (a) the maximum value of the magnetic field and (b) the average intensity of the light there? (c) What is the power of the source? (a) The magnetic field amplitude of the wave is (b) The average intensity is

  11. Radiation pressure This is the force per unit area felt by an object that absorbs light. (Black piece of paper)) This is the force per unit area felt by an object that reflects light backwards. (Aluminum foil)

  12. Polarization of light

  13. 35. In the figure, initially unpolarized light is sent through three polarizing sheets whose polarizing directions make angles of 1 = 40o, 2 = 20o, and 3 = 40o with the direction of the y axis. What percentage of the light’s initial intensity is transmitted by the system? (Hint: Be careful with the angles.) Let Io be the intensity of the unpolarized light that is incident on the first polarizing sheet. The transmitted intensity of is I1 = (1/2)I0, and the direction of polarization of the transmitted light is 1 = 40o counterclockwise from the y axis in the diagram. The polarizing direction of the second sheet is 2 = 20o clockwise from the y axis, so the angle between the direction of polarization that is incident on that sheet and the the polarizing direction of the sheet is 40o + 20o = 60o. The transmitted intensity is and the direction of polarization of the transmitted light is 20o clockwise from the y axis.

  14. 35. In the figure, initially unpolarized light is sent through three polarizing sheets whose polarizing directions make angles of 1 = 40o, 2 = 20o, and 3 = 40o with the direction of the y axis. What percentage of the light’s initial intensity is transmitted by the system? (Hint: Be careful with the angles.) The polarizing direction of the third sheet is 3= 40o counterclockwise from the y axis. Consequently, the angle between the direction of polarization of the light incident on that sheet and the polarizing direction of the sheet is 20o + 40o = 60o. The transmitted intensity is Thus, 3.1% of the light’s initial intensity is transmitted.

  15. Reflection and refraction Snells Law n1 n2

  16. 1 l1 air water L is given by 2 l2 shadow L x 47. In the figure, a 2.00-m-long vertical pole extends from the bottom of a swimming pool to a point 50.0 cm above the water. What is the length of the shadow of the pole on the level bottom of the pool? Consider a ray that grazes the top of the pole, as shown in the diagram below. Here 1 = 35o, l1 = 0.50 m, and l2 = 1.50 m. The length of the shadow is x + L. x is given by x = l1tan1 = (0.50m)tan35o = 0.35 m. According to the law of refraction, n2sin2 = n1sin1. We take n1 = 1 and n2 = 1.33 (from Table 34-1). Then, The length of the shadow is 0.35m + 0.72 m = 1.07 m.

  17. Dispersion: Different wavelengths have different indices of refraction

  18. Total Internal reflection

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