# YDSE with two different mediums

• Saitama
In summary, the distance between the two central maxima is d/4(1+n_1/n_2) apart and the fringe width above and below O is 1.5 times the width of the central maxima.
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.

## 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!

#### Attachments

• YDSE1.png
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• YDSE2.png
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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
... 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".

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Hi Simon Bridge! :)

Simon Bridge said:
...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.

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]

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Sorry for the delay in reply.

Simon Bridge said:
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##.

Pranav-Arora said:
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.]

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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.

#### Attachments

• ydse3.png
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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.

Hi rl.bhat! :)

rl.bhat said:
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?

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.

Pranav-Arora said:
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.

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rl.bhat said:
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

Simon Bridge said:
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?

#### Attachments

• ydse4.png
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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.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?

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## 1. What is YDSE with two different mediums?

YDSE (Young's Double Slit Experiment) with two different mediums refers to a physics experiment that involves light passing through two different materials with different refractive indices. It is used to study the wave-like nature of light and its behavior when passing through different mediums.

## 2. How does YDSE with two different mediums work?

In this experiment, a light source is directed towards two slits, with one slit being covered by a material with a different refractive index compared to the other. The light passing through the slits will diffract and interfere, creating a pattern on a screen placed behind the slits. This pattern can be used to analyze the behavior of light in different mediums.

## 3. What is the significance of YDSE with two different mediums?

This experiment is significant because it helps us understand the wave nature of light and how it behaves when passing through different materials. It also demonstrates the principle of superposition, where two waves can interfere with each other to create a new wave pattern.

## 4. What are some real-life applications of YDSE with two different mediums?

YDSE with two different mediums has practical applications in fields such as optics, spectroscopy, and material science. It can be used to study the properties of different materials and their effect on light, as well as to create interference patterns for various purposes.

## 5. What are the limitations of YDSE with two different mediums?

One limitation of this experiment is that it only works with light waves and cannot be applied to other types of waves. Additionally, the accuracy of the results can be affected by factors such as the quality of the slits and the stability of the light source.

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