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Wize University Physics Textbook (Master) > Wave Optics

Double Slit Diffraction Modified

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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.


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

 dsin⁡θ′=mλ−a(n−1) \boxed{ \ d\sin\theta'=m\lambda-a(n-1) \ }

where aa is the thickness of the material placed in front of the slit.



Wize Concept
Note that the angle θ′\theta' is going to be different from what θ\theta used to be without the different material in front of the slit.






Using the small angle approximation we get:

 y=Dd mλ−Dd a(n−1) \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.


Wize Concept
If the same material is placed in front of both silts (e.g. the experiment is carried underwater), the fringes will be closer together than without the material there (the pattern is squeezed towards the center).






Exam Tip
Don't confuse the mm and the nn in the equation:
  • nn is the index of refraction (specific to each material, always >1>1)
  • mm is the order of interference (always an integer, m=±1,±2, ...m=\pm1, \pm2, \ ...)



Example: Two Media


Two identical light rays (550550 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 1.61.6 while the other still travels in air. What is the minimum thickness of the material?



The speed of light inside the second material is v=cnv=\dfrac{c}{n} , which means it's reduced by a factor of 1.61.6, 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:

Tλ+12=Tλ′\dfrac{T}{\lambda}+\dfrac{1}{2}=\dfrac{T}{\lambda'}


Here TT is the thickness of the material. Use λ′=λn\lambda'=\dfrac{\lambda}{n} for the wavelength inside the material:
Tλ+12=Tnλ\dfrac{T}{\lambda}+\dfrac{1}{2}=\dfrac{Tn}{\lambda}

Rearrange and solve for the thickness TT:

Tnλ−Tλ=12\dfrac{Tn}{\lambda}-\dfrac{T}{\lambda}=\dfrac{1}{2}

T(n−1)λ=12\dfrac{T(n-1)}{\lambda}=\dfrac{1}{2}

T=λ2(n−1)T=\dfrac{\lambda}{2(n-1)}

Plug the numbers is:

=550×10−92(1.6−1)=\dfrac{550\times10^{-9}}{2(1.6-1)}

=4.58×10−7=4.58\times10^{-7} (m)

=458=458 (nm)

Practice: Effective Path Difference


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

a) At what height does the first complete minimum occur?
b) A layer of water (index of refraction 1.331.33) of thickness TT 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?

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.

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 TT. Like with other interference problems, we need to consider the phase difference between the two rays.


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

u1=Asin⁡(kairT+kairΔx1−ωt+ϕ1)u_{1}=A\sin(k_{air}T+k_{air}\Delta x_1-\omega t+\phi_1)

u2=Asin⁡(kwaterT+kairΔx2−ωt+ϕ2)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:

λwater=λairn     and     kwater=nkair\lambda_{water}=\dfrac{\lambda_{air}}{n} \ \ \ \ \ \text{and} \ \ \ \ \ k_{water}=nk_{air}

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


We will say they are initially in sync (that is, ϕ1=ϕ2\phi_1=\phi_2). The phase difference is:


phase difference=\text{phase difference}=

=(kwaterT+kairΔx2−ωt+ϕ2)−(kairT+kairΔx1−ωt+ϕ1)=(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})

=(kwaterT+kairΔx2)−(kairT+kairΔx1)=(k_{water}T+k_{air}\Delta x_2)-(k_{air}T+k_{air}\Delta x_1) \\

=T(kwater−kair)+kair(Δx2−Δx1)=T(k_{water}-k_{air})+k_{air}(\Delta x_2-\Delta x_1) \\

=T(2πλwater−2πλ)+2πλ(Δx2−Δx1)= T\bigg(\dfrac{2\pi}{\lambda_{water}}-\dfrac{2\pi}{\lambda}\bigg)+\dfrac{2\pi}{\lambda} (\Delta x_2-\Delta x_1) \\

=T(2πnλ−2πλ)+2πλ(Δx2−Δx1)= T\bigg(\dfrac{2\pi n}{\lambda}-\dfrac{2\pi}{\lambda}\bigg)+\dfrac{2\pi}{\lambda} (\Delta x_2-\Delta x_1)

 =2πλ[T(n−1)+dsin⁡θ′] \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 2π2\pi (we write 2πm2\pi m):

2πm=2πλ[T(n−1)+dsin⁡θ′]2\pi m = \dfrac{2\pi}{\lambda}\bigg[T(n-1)+ d\sin\theta'\bigg]

 mλ=T(n−1)+dsin⁡θ′ \boxed{ \ m \lambda = T(n-1)+ d\sin\theta' \ }


Exam Tip
Don't confuse the mm and the nn in this equation:
  • nn is the index of refraction (specific to each material, always >1>1)
  • mm is the order of interference (always an integer, m=±1,±2, ...m=\pm1, \pm2, \ ...)


Method 2: Speed of light reduced in a material


If light travels through water, its speed decreases to:

vwater=cnv_{water}=\dfrac{c}{n{}}

where nn 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 Δt\Delta t that it takes the light to go through the water thickness TT, the light travels a distance cΔtc \Delta t:

Δt=twater−tair\Delta t=t_{water}-t_{air}

Δt=Tv−Tc\Delta t=\dfrac{T}{v}-\dfrac{T}{c}

Δt=Tc/n−Tc\Delta t=\dfrac{T}{c/n}-\dfrac{T}{c}

Δt=Tc(n−1)\Delta t=\dfrac{T}{c}(n-1)

cΔt=T(n−1)c\Delta t=T(n-1)


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

mλ=dsin⁡θ′+cΔtm\lambda=d\sin\theta'+c\Delta t

 mλ=dsin⁡θ′+T(n−1) \boxed{ \ m\lambda=d\sin\theta'+T(n-1) \ }

This is the new effective path difference for the problem.


Wize Concept
Note that the angle is now going to be different. Still very small, but different.