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YDSE with two different mediums

  1. Feb 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the scheme of YDSE as shown in fig filled with transparent of refractive indices ##n_1## and ##n_2## respectively.The slits are at a distance d apart. Find:

    1. Location and width of central maxima.

    2. Condition for constructive and destructive inteference

    3. Fringe width above and below O.


    2. Relevant equations



    3. The attempt at a solution
    I have never encountered such kind of setup for YDSE so I am not sure if the following is correct.

    Consider the rays shown in attachment 2. I am thinking that if the refracted ray is extended backwards, it meets the cardboard with slits at point P. I can consider point P as the "effective" second slit. So I need the distance between the slit ##S_1## and this point i.e (x+d/2).

    From Snell's law and trigonometry, I found ##x=\frac{n_1d}{2n_2}##. Hence, the central maxima would lie on the perpendicular bisector of line joining ##S_1## and P i.e at a distance of ##d/4(1+n_1/n_2)##. The central maxima is at distance ##(d/4)(1-n_1/n_2)## from O. Is this correct?

    Any help is appreciated. Thanks!
     

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  2. jcsd
  3. Feb 27, 2014 #2

    Simon Bridge

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    The effect of the media is to change the effective path length... there have been treatments of this:
    http://www.physics.ohio-state.edu/~gohlke/pedagogy/Phys133I_Diffraction.pdf [Broken]
    ... just puts a glass slide after one slit.
    Which is more usual.

    For the case that the media goes all the way to the screen - say a glass block on one side and some liquid on the other - you still need the whole path, because the phase at P is also important and P is also in different places for different positions along the screen. i.e. you cannot treat point P as an effective "other slit".
     
    Last edited by a moderator: May 6, 2017
  4. Feb 28, 2014 #3
    Hi Simon Bridge! :)

    I am not sure if I agree with this but the value of ##x## comes out to be the same irrespective of the position on the screen. :confused:
     
  5. Feb 28, 2014 #4

    Simon Bridge

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    OK, so lets label a few more bits of your diagram so I can talk about them:
    Let the point of refraction be ##Q## and the point on the screen be ##R##, the distance ##|OR|=z##.

    sanity check:

    for ##d<<D##, ##\alpha \simeq \pi/2##, which makes ##\beta \simeq \sin^{-1}\frac{n_2}{n_1}##
    ... which fixes the value of ##\beta## to a narrow range of possible angles: approximately constant.

    Given constant ##\beta##, the position of R will determine the position of Q and P.
    You can check this with a straight edge.

    [edited for brevity]
     
    Last edited: Mar 1, 2014
  6. Mar 1, 2014 #5
    Sorry for the delay in reply. :redface:

    I am in full agreement that position of Q depends on position of R but I don't see how the position of P gets affected. Can you please show me what's wrong with the calculations I have done in my attempt? Please tell me if anything is unclear in my attempt, I will post the steps for how I arrived at ##x##.
     
  7. Mar 1, 2014 #6

    Simon Bridge

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    Maybe I misunderstood: on your diagram, P Q and R are on the same line. You can demonstrate the dependence of P on R by using a straight edge - same slope (because β is a constant), different values of R. Or you can keep P fixed, change R, and notice that the angle β has to change quite a lot to accommodate this, which it cannot do if D>>d. Unless you see something I missed ;)

    You have not shown your working so I cannot, besides, it is not my practice to check someone else's working for them - however:

    You have concluded that: ##x=\frac{n_1d}{2n_2}##
    ... which looks the same as saying that ##\frac{d}{2}=x\sin\beta##
    ... does this make sense geometrically?

    I don't see how you got from the diagram you provided to this relation.
    (BTW: we may want to label the point half way between S1 and S2 as O' or something.)

    ... never mind - I think I see.
    I'll get back to you.

    [edit: back - I have a feeling about what you did but I would like you to show me before I make wild guesses.]
     
    Last edited: Mar 1, 2014
  8. Mar 2, 2014 #7
    I post my working. :)

    I have marked Q as the point of refraction and O' as the point lying exactly midway between ##S_1## and ##S_2##.

    In ##\Delta S_2O'Q##,
    $$\tan(\pi/2-\alpha)=\frac{d/2}{y}=\cot\alpha=\frac{1}{\tan\alpha}$$
    Since the angles are small, I can write ##\tan\alpha \approx \alpha## i.e
    $$\frac{1}{\alpha}=\frac{d}{2y}\Rightarrow \alpha=\frac{2y}{d}\,\,\,\, (*)$$
    From Snell's law: ##n_2\sin\alpha=n_1\sin\beta \Rightarrow n_2\alpha=n_1\beta \Rightarrow \beta=(n_2/n_1)\alpha##
    In ##\Delta PO'Q##,
    $$\tan(\pi/2-\beta)=\frac{x}{y}=\frac{1}{\tan\beta}$$
    Again ##\tan\beta \approx \beta##, i.e
    $$\frac{1}{\beta}=\frac{x}{y} \Rightarrow \beta=\frac{y}{x} \Rightarrow \frac{n_2}{n_1}\alpha=\frac{y}{x}\,\,\, (**)$$
    Dividing (*) and (**), I get
    $$\frac{\alpha}{(n_2/n_1)\alpha}=\frac{2y/d}{y/x}\Rightarrow \frac{n_1}{n_2}=\frac{2x}{d} \Rightarrow x=\frac{n_1}{n_2}\frac{d}{2}$$
    ...which is the same relation I posted in my attempt.

    Please let me know if anything is still unclear.
     

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  9. Mar 2, 2014 #8

    rl.bhat

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    Hi Pranav, in your derivation you have assumed α is very small. If α is the angle made by the incident ray with the normal, it will be small only when y is very small. In that case the screen should be very close the the slits.
     
  10. Mar 2, 2014 #9
    Hi rl.bhat! :)

    Ah yes, I see it now, Simon Bridge pointed this out as ##\alpha \approx \pi/2##, I should have been careful. :redface:

    But now how should I approach this problem? :confused:
     
  11. Mar 2, 2014 #10

    Simon Bridge

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    That was the feeling I had - I could only get your result by assuming that ##\tan\alpha \approx \sin\alpha## etc. The "small angles" approximation for D>>d is for the angle θ=∠RO'O.

    The impact of D>>d on α and β is that ##\beta = \sin^{-1}(\frac{n_2}{n_1})## ... as per post #4.

    Approach:
    Use the phasor description - work out the phase of the ray from S1 and S2 as a function of z.
    The amplitude at R will be proportional to the vector sum of the two phasors.

    There is a reason you don't find this case discussed much online: it is not easy.
     
  12. Mar 2, 2014 #11

    rl.bhat

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    The optical path difference between the two slits and the central maximum is given by ( μ1 - μ2 )t where t = ( D^2 + d^2 )^1/2
    In the two region the wavelengths are different.
     
    Last edited: Mar 2, 2014
  13. Mar 2, 2014 #12

    Simon Bridge

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    Maybe - but that sentence does not make grammatical sense and the equation seems odd.
    You may need to explain the reasoning behind it and be a bit clearer with how it is related to the OPD.

    But the frequencies are the same - hence the phase approach.

    Note: Each ray spends different amounts of time in each medium.
    http://en.wikipedia.org/wiki/Optical_path_length
     
  14. Mar 3, 2014 #13
    I am not sure but do I have to work out the path difference? :confused:

    $$\Delta x=n_1(S_1R-QR)-n_2S_2Q$$
    where ##\Delta x## is the path difference.
    $$S_1R=\sqrt{\left(z-\frac{d}{2}\right)^2+D^2}$$
    $$\sin\left(\frac{\pi}{2}-\beta\right)=\frac{z}{QR}\Rightarrow QR=\frac{z}{\cos\beta}=\frac{n_1z}{\sqrt{n_1^2-n_2^2}}$$
    $$S_2Q \approx O'Q=D-QO=D-z\tan\beta=D-\frac{zn_2}{\sqrt{n_1^2-n_2^2}}$$
    I think I need to approximate ##S_1R## but how? :confused:
     

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    Last edited: Mar 3, 2014
  15. Mar 3, 2014 #14

    Simon Bridge

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    Keeping to the same labels as before:
    You want to divide the screen into two areas - z>0 and z<0.
    You are currently working for z>0

    iirc: the optical path difference (OPD) should be a whole number of vacuum wavelengths.
    http://www.physics.gla.ac.uk/~johannes/optics_lecture/overheads11_2_2_6-8.pdf [Broken]
    http://www.ph.utexas.edu/~asimha/PHY315/OpticalPathLength.pdf [Broken]

    Did you notice: $$\frac{D-y}{\sqrt{(D-y)^2+z^2}}\simeq \frac{n_2}{n_1}$$

    Note: in the regular setup, you get to exploit ##\theta_1=\theta_2=\theta## for D>>d, and ##D\tan\theta=z##.

    In this case ##\theta_1 = \theta \neq \theta_2##

    So you may get away with approximating ##|S_1R|=\sqrt{z^2+D^2}##.

    I don't really know, I havn't done it.
    You will have a lot of messing about before you hit one something useful.
    Try working it out without the approximations to get a very messy ##\Delta\phi = f(z):z>0##, then plot phase-change vs z for some handy values. See what happens.

    note: $$\frac{1}{\tan\beta} = \sqrt{\left(\frac{n_1}{n_2}\right)^2-1}$$ ... this number appears a LOT and it is a constant for the setup - so maybe give it it's own letter?
     
    Last edited by a moderator: May 6, 2017
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