Putting a slab of some material just after the slits in Young's model

[Adesh]

Homework Statement

Find the equations of path difference and the equations for maxima and minima.

Relevant Equations

#path~difference ~= d\sin\theta#.

If I put a slab of some material whose refractive index is $\mu$ and width is $D$, in front of slits in Young's Double Slit model, then

In the figure you can see that I have placed the slab just after the slits. So, when rays going to come out of the slab they will bend away from normal (here I'm assuming that the material of the slab is optically denser than the air) and due to this bend they will intersect at P' and if the slab were to be absent they would have met at P. So, a pattern at P has shifted to P' but what is the itexematics of it? How to derive equations for path difference in this situation? Do we draw a perpendicular from $S_1$ to $S_2P$ like usual, but $S_2P$ is not a single line it gets bent just after the slab. How should I begin?

 

[ehild]

Homework Statement:: Find the equations of path difference and the equations for maxima and minima.
Homework Equations:: #path~difference ~= d\sin\theta#.

If I put a slab of some material whose refractive index is $\mu$ and width is $D$, in front of slits in Young's Double Slit model, then

View attachment 255168
In the figure you can see that I have placed the slab just after the slits. So, when rays going to come out of the slab they will bend away from normal (here I'm assuming that the material of the slab is optically denser than the air) and due to this bend they will intersect at P' and if the slab were to be absent they would have met at P. So, a pattern at P has shifted to P' but what is the itexematics of it? How to derive equations for path difference in this situation? Do we draw a perpendicular from $S_1$ to $S_2P$ like usual, but $S_2P$ is not a single line it gets bent just after the slab. How should I begin?

You have to determine the optical path difference between the rays separately in the slab and in air, and then add up the differences.
The optical path-length is the physical length multiplied by the refractive index.
You can consider the interfering waves parallel, as the screen is far away, and the distance between the slits is very small. No need to calculate with two angles of incidence. The parallel waves become a single wave.

 

[Adesh]

You have to determine the optical path difference between the rays separately in the slab and in air, and then add up the differences.
The optical path-length is the physical length multiplied by the refractive index.
You can consider the interfering waves parallel, as the screen is far away, and the distance between the slits is very small. No need to calculate with two angles of incidence. The parallel waves become a single wave.

In your short answer you have given me so much information, thank you for that.

Should I do that perpendicular thing one time in the slab and then in the air? I really want to know why we have to use optical path length in this case, why does geometric path length (the simple case from where we derive $path ~difference ~= d\sin\theta$) doesn’t work here?

 

[ehild]

The interference of two parallel traveling waves depend on the phase difference between them. The speed of light in a medium depends on the refractive index, so the wavelength is different in different media.
If the wavelength becomes shorter, the same distance traveled causes greater phase change.
What is the Mathematical form of a wave traveling in the direction x?

 

[Adesh]

If the wavelength becomes shorter, the same distance traveled causes greater phase change.

Does this conclusion came from $$ phase ~difference ~= \frac{2\pi}{\lambda} x$$
Where $x$ is the path difference? I’m not able to think intuitively about it. I understand that when $\lambda$ decreases the value of the expression increases (because when denominator decrease the value of the expression gets bigger) but how to think of it intuitively.

What is the Mathematical form of a wave traveling in the direction x?

$y(x,t) = A \cos (\omega t - kx)$.

 

[ehild]

Does this conclusion came from $$ phase ~difference ~= \frac{2\pi}{\lambda} x$$
Where $x$ is the path difference?

Yes. And how does the wavelength in a medium depend on the refractive index?

I’m not able to think intuitively about it. I understand that when $\lambda$ decreases the value of the expression increases (because when denominator decrease the value of the expression gets bigger) but how to think of it intuitively.

$y(x,t) = A \cos (\omega t - kx)$.

The "k" in the formula is the "wavenumber" 2pi/lambda.
You know that a wave is periodic both in time and place. If the time period is T, the wave travels $\lambda=vT$ distance in one period. V= c/n where c is the speed of light in vacuum and n is the refractive index. The refractive index in materials is usually greater then in vacuum where it is 1, so the wavelength is shorter in a medium than in vacuum.

 

[Adesh]

Yes. And how does the wavelength in a medium depend on the refractive index?

The "k" in the formula is the "wavenumber" 2pi/lambda.
You know that a wave is periodic both in time and place. If the time period is T, the wave travels $\lambda=vT$ distance in one period. V= c/n where c is the speed of light in vacuum and n is the refractive index. The refractive index in materials is usually greater then in vacuum where it is 1, so the wavelength is shorter in a medium than in vacuum.

So, that means the wave will travel a shorter distance in time $T$ because speed is reduced. Therefore, the distance traveled in time $T$ is (when a medium of refractive index n is placed) $$ \lambda_n = v_n T$$.
Am I right? What to do next?

 

[ehild]

So, that means the wave will travel a shorter distance in time $T$ because speed is reduced. Therefore, the distance traveled in time $T$ is (when a medium of refractive index n is placed) $$ \lambda_n = v_n T$$.
Am I right? What to do next?

How is the wavelength in the medium related to the vacuum wavelength? Did you mean, that it is the vacuum wavelength divided by the refractive index?
You wrote that the phase difference is $\frac{2\pi}{\lambda} x$
what is equal to $\frac{2\pi}{\lambda_0/n} x=\frac{2\pi}{\lambda_0 }nx$ as if the distance traveled was multiplied by n.
Assume an angle of incidence, determine the geometric path difference in the slab, if there is any, multiply it by the refractive index. Add the path difference arising in the vacuum, outside the slab.

 

[Adesh]

In this diagram I have drawn the rays parallel because it would make geometry easier to see.

I have drawn perpendiculars $S_1A$ and $S'_1A'$ on the rays $S_2A$ and $S'_2A'$. By simple geometry angle $S_2 S_1A = \theta_1$ and $S'_2 S'_1 A' =\theta_2$ . Now, by trigonometry $S_2 A = d \sin\theta_1$ and $S'_2 A' = d\sin\theta_2$. Therefore, $$ total~path~difference~= d\sin\theta_1 + d\sin\theta_2$$. Am I correct so far?

 

[ehild]

In this diagram I have drawn the rays parallel because it would make geometry easier to see.

View attachment 255171

There is also a path difference between the rays at the exit from the slab.

 

[Adesh]

There is also a path difference between the rays at the exit from the slab.
View attachment 255174

Does that mean the two rays were of equal length inside the slab? So, total path difference is $d\sin\theta_2$ ?

 

[ehild]

Does that mean the two rays were of equal length inside the slab? So, total path difference is $d\sin\theta_2$ ?

Yes.

 

[Adesh]

Yes.

By Snell’s Law $$ n_1 \sin\theta_1 = n_2\sin\theta_2 \\
\sin\theta_2 = \frac{n_1}{n_2} \sin\theta_1$$
Therefore, the path difference is $d ~\frac{n_1}{n_2} ~\sin\theta_1$. So, for constructive interference we have $$ d~\frac{n_1}{n_2} ~ \sin\theta_1 = m\lambda$$ where $m$ is any integer and $\lambda$ is wavelength of the wave in medium with refractive index $n_1$ (that is air in our case).

So, why does the thickness of the slab matters? This equation is totally independent of width of the slab. Please explain.

 

[ehild]

So, why does the thickness of the slab matters? This equation is totally independent of width of the slab. Please explain.

You got the condition for constructive interference in terms of the angles - theta1 or theta2. But you observe the light spots on a screen. At what distance x is the first spot from the central one? Assume the slab has thickness T and the screen is at distance L from the slits.
What happens if you observe the spots on the rare side of the slab (which is matted)?

 

[Adesh]

You got the condition for constructive interference in terms of the angles - theta1 or theta2. But you observe the light spots on a screen. At what distance x is the first spot from the central one? Assume the slab has thickness T and the screen is at distance L from the slits.
What happens if you observe the spots on the rare side of the slab (which is matted)?

Okay. So, when we solve for shifting we need to consider the thickness, I got it.
Thank you so much. I learned very much today from you.