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Polarization, Double Slit, Single Slit

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data

    This is an old test question that I failed, but this material will resurface for the final so I just wanted a detailed description of what the full answer would've been. I'll include the rubric too so you'll know what I'm up against. Can you include explanations and pictures? I promise I already attempted it, and FAILED in class. I just don't have a scanner to scan you the exam.
    Consider a 0.5mm diameter beam of light that has a wavelength of λ=10nm that is an equal mixture of horizontal and vertical polarizations. We pass this beam through a vertically oriented double-slit that also acts as a polarizer. The amplitude of the electromagnetic field that produces this beam is 100 N/C. Find the average intensity of this beam both before and after it passes through the two slits. Explain your reasoning very briefly. If we place a screen 10 cm from the slits, what is the distance between adjacent (double slit) bright spots? What is the distance between adjacent (single slit ) dark fringes? What is the power of the beam incident on the screen? How many photons per second hit the screen? Suppose we replaced the light beam with a 100eV beam of electrons. Explain what happens to the interference patterns when we do this.
    a= .025mm d=.04mm
    2pts for a correct equation leading to a correct answer for the average intensity of the before it reaches the slits
    2pts for a correct value for the average intensity of the beam after it passes through the slits and briefly noting how you arrived at this
    4pts for correct algebra and geometry leading to a correct algebraic equation for the distance between adjacent (double-slit) bright spots in terms of the distance to the screen, the slit separations, and the wavelength of the beam
    1pt for the correct value for this separation
    2pt for the correct equation for the distance between adjacent (single slit) dark fringes in terms of the distance to the screen, the slit width, and the wavelength of the beam
    1pt for correct value of this separation
    4pts for correctly noting the relationship between intensity and power and using this to derive an expression for the power in terms of the intensity and diameter of the beam
    1pt for the correct value for the beam’s power incident on the screen
    2pts for correctly noting the relationship between energy of the individual photons and their wavelength, and using this to derive an expression for the number of photons per second that impinge on the screen
    1pt for the correct value for the number of photons that hit the screen
    3pts for correctly determining the wavelength of the electron beam
    2pts for correctly explaining what happens to the adjacent bright spots and dark fringes when we make this change



    2. Relevant equations

    k/c=30kg*m2/C2
    tan theta=sintheta=theta for a small theta
    rest energy/mass of an electron: mc2= 511000eV
    theta=inverse sin of (nλ/a)
    theta = inverse sin of (nλ/d)
    3. The attempt at a solution

    I found the intensity using I= cE2/8∏K, but I was wrong. I also said that becasue light is polarized the average intensity after will be the exact opposite of the intensity before, that is also wrong.
    I used the wrong equation for the double slit portion. which then carried out for the majority of the problem.I also used the incorrect value for the area to find power, and I'm not sure why that was wrong-->.04mm
     
  2. jcsd
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