Wize University Physics Textbook (Master) > Wave Optics

Double Slit Diffraction Modified

Effective Path Difference

Sometimes you will encounter problems where there are materials other than air in a two-slit experiment. For example, a piece of glass may be placed in front of one of the slits, or a layer of water, etc.

Image illustrating Effective Path Difference

The the new effective path difference for the problem is given by:

\boxed{ \ d\sin\theta'=m\lambda-a(n-1) \ }

where $a$ is the thickness of the material placed in front of the slit.

Using the small angle approximation we get:

\boxed{ \ y=\dfrac{D}{d} \ m\lambda-\dfrac{D}{d} \ a(n-1) \ }

This means that the distance on the screen is now smaller than what it used to be without the material added in front of the slit, and the pattern in shifted downwards.

Example: Two Media

Two identical light rays ( $550$ nm) are separated and then combined to interfere destructively. While they are separated, one of the rays travels through a material with index of refraction $1.6$ while the other still travels in air. What is the minimum thickness of the material?

$Image illustrating Example: Double Slit Diffraction$

The speed of light inside the second material is $v=\dfrac{c}{n}$ , which means it's reduced by a factor of $1.6$ , and there will be more wavelengths inside that material than through air, over the same distance.

The minimum thickness occurs when exactly half a wavelength more is being fit inside the material compared to the light ray outside:

$\dfrac{T}{\lambda}+\dfrac{1}{2}=\dfrac{T}{\lambda'}$

Here $T$ is the thickness of the material. Use $\lambda'=\dfrac{\lambda}{n}$ for the wavelength inside the material:

$\dfrac{T}{\lambda}+\dfrac{1}{2}=\dfrac{Tn}{\lambda}$

Rearrange and solve for the thickness $T$ :

$\dfrac{Tn}{\lambda}-\dfrac{T}{\lambda}=\dfrac{1}{2}$

$\dfrac{T(n-1)}{\lambda}=\dfrac{1}{2}$

$T=\dfrac{\lambda}{2(n-1)}$

Plug the numbers is:

$=\dfrac{550\times10^{-9}}{2(1.6-1)}$

$=4.58\times10^{-7}$ (m)

$=458$ (nm)

Practice: Effective Path Difference

Consider a two-slit interference problem with distance between slits $d=0.10$ mm and distance to the screen $D=5.0$ m. Assume $y\ll D$ , that is, $\theta$ is very small. We shine a blue laser at the slits ( $\lambda=450$ nm).

a) At what height does the first complete minimum occur?

b) A layer of water (index of refraction $1.33$ ) of thickness $T$ is placed in front of the bottom slit. What is the smallest thickness of water that will have no impact on where we will find maxima in the interference pattern?

a) Height of 1st min (answer in mm, 2 decimal places, don't include units):b) Thickness (answer in $\mu m$, 2 decimal places, don't include units):

I don't know

Effective Path Difference

There are two equivalent methods to find the new effective path difference.

Method 1: Phase differences

Consider a setup where the bottom slit is covered by a layer of water of thickness $T$ . Like with other interference problems, we need to consider the phase difference between the two rays.

Let's use the wave equation $u=A\sin(k\Delta x-\omega t)$ for each of our two waves:

$u_{1}=A\sin(k_{air}T+k_{air}\Delta x_1-\omega t+\phi_1)$

$u_{2}=A\sin(k_{water}T+k_{air}\Delta x_2-\omega t+\phi_2)$

The wavelength and wave number of light will be modified in water:

$\lambda_{water}=\dfrac{\lambda_{air}}{n} \ \ \ \ \ \text{and} \ \ \ \ \ k_{water}=nk_{air}$

where $n$ is the index of refraction of water (or any other material).

We will say they are initially in sync (that is, $\phi_1=\phi_2$ ). The phase difference is:

$\text{phase difference}=$

$=(k_{water}T+k_{air}\Delta x_2-\omega t+\cancel{\phi_2})-(k_{air}T+k_{air}\Delta x_1-\omega t+\cancel{\phi_1})$

$=(k_{water}T+k_{air}\Delta x_2)-(k_{air}T+k_{air}\Delta x_1) \\$

$=T(k_{water}-k_{air})+k_{air}(\Delta x_2-\Delta x_1) \\$

$= T\bigg(\dfrac{2\pi}{\lambda_{water}}-\dfrac{2\pi}{\lambda}\bigg)+\dfrac{2\pi}{\lambda} (\Delta x_2-\Delta x_1) \\$

$= T\bigg(\dfrac{2\pi n}{\lambda}-\dfrac{2\pi}{\lambda}\bigg)+\dfrac{2\pi}{\lambda} (\Delta x_2-\Delta x_1)$

$\boxed{ \ = \dfrac{2\pi}{\lambda}\bigg[T(n-1)+ d\sin\theta'\bigg] \ }$

For example, if we apply this when looking at constructive interference, we need the phase angle to be a multiple of $2\pi$ (we write $2\pi m$ ):

$2\pi m = \dfrac{2\pi}{\lambda}\bigg[T(n-1)+ d\sin\theta'\bigg]$

$\boxed{ \ m \lambda = T(n-1)+ d\sin\theta' \ }$

Method 2: Speed of light reduced in a material

If light travels through water, its speed decreases to:

$v_{water}=\dfrac{c}{n{}}$

where $n$ is the index of refraction of water (or any other material).

We need to consider how much further ahead the light from the top slit got while the bottom-slit-light was going through water. In the extra time $\Delta t$ that it takes the light to go through the water thickness $T$ , the light travels a distance $c \Delta t$ :

$\Delta t=t_{water}-t_{air}$

$\Delta t=\dfrac{T}{v}-\dfrac{T}{c}$

$\Delta t=\dfrac{T}{c/n}-\dfrac{T}{c}$

$\Delta t=\dfrac{T}{c}(n-1)$

$c\Delta t=T(n-1)$

Again, for constructive interference in air we have $m\lambda=d\sin\theta$ . How does this equation change with a layer of water? We need to add the extra $c\Delta t$ distance traveled by light to the path difference:

$m\lambda=d\sin\theta'+c\Delta t$

$\boxed{ \ m\lambda=d\sin\theta'+T(n-1) \ }$

This is the new effective path difference for the problem.