YDSE with two different mediums

[Saitama]

Homework Statement

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.

Homework Equations

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!

 

[Simon Bridge]

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
... 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".

 

[Saitama]

Hi Simon Bridge! :)

...and P is also in different places for different positions along the screen.

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.

 

[Simon Bridge]

OK, so let's 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]

 

[Saitama]

Sorry for the delay in reply.

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

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 $x$.

 

[Simon Bridge]

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.

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 ;)

Can you please show me what's wrong with the calculations I have done in my attempt?

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.]

 

[Saitama]

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.

 

[rl.bhat]

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.

 

[Saitama]

Hi rl.bhat! :)

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.

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

But now how should I approach this problem?

 

[Simon Bridge]

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.

 

[rl.bhat]

Hi rl.bhat! :)
Ah yes, I see it now, Simon Bridge pointed this out as $\alpha \approx \pi/2$, I should have been careful.

But now how should I approach this problem?

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.

 

[Simon Bridge]

The optical path difference the the two slits and the central maximum is given by ( μ1 - μ2 )t where t = ( D^2 + d^2 )^1/2

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.

In the two region the wavelengths are different.

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

 

[Saitama]

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.

I am not sure but do I have to work out the path difference?

$$\Delta x=n_1(S_1R-QR)-n_2S_2Q$$
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?

 

[Simon Bridge]

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
http://www.ph.utexas.edu/~asimha/PHY315/OpticalPathLength.pdf

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?