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Three slit minima and maxima and path differences

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Find at which angles θ the interference picture that appears on a distant screen
    made by three thin slits separated by distance d and enlightened by a source of wavelength λ (see figure)

    a) Shows its maxima.
    b) Shows its minima.
    26Jlomz.png
    multiple slit diffraction, d, ##\lambda##, ##\theta##
    2. Relevant equations
    $$dsin\theta=m\lambda$$
    $$dsin\theta=(m\lambda)/3$$
    $$E(t)=E_0cos(k(r_2-r_1)-\omega(t))$$
    $$t=top,m=middle,b=bottom$$

    3. The attempt at a solution
    $$E_t=E_0cos(k(r_t-r_b)-\omega(t))$$
    $$E_m=E_0cos(k(r_t-r_m)-\omega(t))$$
    $$E_b=E_0cos(-\omega(t))$$

    However, the solutions gives me something like this.
    Oz1l7od.png
    My thought process was that the top ray travels the most or has the biggest path difference. I don't get why the solutions give me these equations. Or am I getting my path differences mixed up? Thanks for the help.
     
  2. jcsd
  3. Nov 15, 2016 #2

    Charles Link

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    I don't know how advanced your textbook is, but if you have the formula for the interference pattern for ## N ## slits (It works for any integer N that is 2 or greater): ## I(\theta)=I_o \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2)} ## where ## \phi= (\frac{2 \pi}{\lambda}) d \sin{\theta} ##, you can use it to solve for ## I(\theta) ## for the case where ## N=3 ##. ## \\ ## The zero's of the function ## I(\theta) ## occur when the numerator is equal to zero, except when the denominator is also zero. When the denominator is also zero, along with the numerator, you get a maximum and ## I(\theta_{max})=I_o N^2 ##.(You can take the limit where ## \theta ## approaches that angle. At that precise angle, the expression is undefined, but assumed to be the same value as the limit.) ## \\ ## I will let you try to work out this result assuming you are allowed to use this formula. It's a simple matter to determine the angles ## \theta ## for which the numerator and/or the denominator is equal to zero. ## \\ ## Note: If you wanted to work with sinusoidal ## E ## fields, instead of simply using the formula for ## I(\theta) ##, ## \phi ## would be the phase of the middle slit, with the top slit at zero (relative phase), and the bottom slit at ## 2 \phi ##. The algebra for the case of just 3 sources isn't terribly complex (using trigonometric identities to compute the sums, etc.), but it is really easier to just use the general formula given above.
     
    Last edited: Nov 15, 2016
  4. Nov 16, 2016 #3

    Charles Link

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    One additional comment is the formula for the intensity for N slits is derived by first summing sinusoidal E fields using complex variables (## exp(i \omega t) ## etc.,) and summing the geometric series. For N=2 and maybe for N=3, it is possible to do this calculation using ## cos( \omega t) ##, but for anything larger, it is much easier to do this calculation with complex variables.
     
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