Thomas Young's double slit experiment

[Physics345]

Homework Statement

a) Explain why a pattern of bright and dark fringes visible on a screen when a light is shone through a double slit.

b) Upon using Thomas Young's double-slit experiment to obtain measurements, the following data were obtained. use this data to determine the wavelength of light being used to create the interference pattern. do this in three different ways.

Homework Equations

mλ=d sin⁡θ
mλ=dxm/L
∆x=Lλ/d

The Attempt at a Solution

a)
When the wave encounters a barrier such as a slit it spreads out into two dimensions causing diffraction, when using a double slit there is going to be two waves that will overlap, since the light passes through two barriers (the double slits). When the overlapping occurs, there will be regions where they overlap constructively. Constructive interference occurs when peaks line up over peaks or valleys over valleys causing waves that are in phase, creating the bright fringes on the screen in the directions of the constructive interference. Where they overlap destructively you get a dark fringe. Destructive interference on the other hand occurs when peaks match up with the valleys and in between them there is a destructive point, creating dark fringes on the screen in the directions of the destructive interference.

https://www.khanacademy.org/science...e-of-light-waves/v/youngs-double-split-part-1

b)
b)
θ=1.12°
m=8
L=302cm=3.02m
4∆x=2.95 / 4
∆x=0.7375 cm=0.0007375m
∆x=7.375 × 10^-4 m
x_4=2.95 × 10^-2 m
d=2.5 ×1 0^-4 m

Method 1:
mλ=d sin⁡θ
λ=(d sin⁡θ)/m
λ=(2.5×10^-4)(sin⁡1.12°) / 8
λ=6.11×10^-7
λ=611 nm

Method 2: do calculations here
mλ=dxm/L
λ=dxm/mL
λ=(2.95×10^-2)(7.375×10^-4) / (4(3.02))
λ=6.11×10^-7
λ=611 nm

Method 3:
∆x=Lλ/d
λ=d∆x/L
λ=(2.5×10^-4 )(7.375×10^-4 ) / 3.02
λ=6.11×10^-7
λ=611 nm
Therefore the wavelength of light being used is 611 nm

 

[Gene Naden]

Can you please make a sketch showing the geometry of the experiment? I think that will help us.

 

[Physics345]

Can you please make a sketch showing the geometry of the experiment? I think that will help us.

Well that is clearly covered in the video I linked, it would be pointless for me to just do the same thing he did. I'll do it but I honestly do not see the merit in it, when the question does not ask for it. I'm not trying to be condescending here, I just want to understand why.

 

[Gene Naden]

The reason for looking at the sketch is that the correct application of the formulas depends on the geometrical relatioinships involved. Just plugging in formulas would not be very meaningful. The sketch should show the relationships between the inter-fringe distance, the angles, the wavelength, etc. There is no way to understand a double-slit experiment without looking at the geometry involved.

 

[Physics345]

The reason for looking at the sketch is that the correct application of the formulas depends on the geometrical relatioinships involved. Just plugging in formulas would not be very meaningful. The sketch should show the relationships between the inter-fringe distance, the angles, the wavelength, etc. There is no way to understand a double-slit experiment without looking at the geometry involved.

The lesson has never asked us to do that nor has it done it in it's examples. seems more advanced then the scope of my course. Regardless I will get on it right after I finish typing some work out.

 

[Physics345]

Okay I'm working on the diagram now, also was my math wrong?

Edit: After thinking for a while I can't figure out how to do this my textbook is not showing such an example I have no clue where to start.

 

[Physics345]

I just realized I didn't even state the given let me re write the question.

Homework Statement

a) Explain why a pattern of bright and dark fringes visible on a screen when a light is shone through a double slit.

b) Upon using Thomas Young's double-slit experiment to obtain measurements, the following data were obtained. use this data to determine the wavelength of light being used to create the interference pattern. do this in three different ways.
- The angle to the eigth maximum is 1.12 degrees.
- The distance from the slits to the screen is 302 cm
- The distance from the first minimum to the fifth minimum is 2.95 cm
- The distance between the slits is 0.00025 m.

Homework Equations

mλ=d sin⁡θ
mλ=dxm/L
∆x=Lλ/d

The Attempt at a Solution

a)
When the wave encounters a barrier such as a slit it spreads out into two dimensions causing diffraction, when using a double slit there is going to be two waves that will overlap, since the light passes through two barriers (the double slits). When the overlapping occurs, there will be regions where they overlap constructively. Constructive interference occurs when peaks line up over peaks or valleys over valleys causing waves that are in phase, creating the bright fringes on the screen in the directions of the constructive interference. Where they overlap destructively you get a dark fringe. Destructive interference on the other hand occurs when peaks match up with the valleys and in between them there is a destructive point, creating dark fringes on the screen in the directions of the destructive interference.

https://www.khanacademy.org/science...e-of-light-waves/v/youngs-double-split-part-1

b)
b)
θ=1.12°
m=8
L=302cm=3.02m
4∆x=2.95 / 4
∆x=0.7375 cm=0.0007375m
∆x=7.375 × 10^-4 m
x_4=2.95 × 10^-2 m
d=2.5 ×1 0^-4 m

Method 1:
mλ=d sin⁡θ
λ=(d sin⁡θ)/m
λ=(2.5×10^-4)(sin⁡1.12°) / 8
λ=6.11×10^-7
λ=611 nm

Method 2: do calculations here
mλ=dxm/L
λ=dxm/mL
λ=(2.95×10^-2)(7.375×10^-4) / (4(3.02))
λ=6.11×10^-7
λ=611 nm

Method 3:
∆x=Lλ/d
λ=d∆x/L
λ=(2.5×10^-4 )(7.375×10^-4 ) / 3.02
λ=6.11×10^-7
λ=611 nm
Therefore the wavelength of light being used is 611 nm