# Thomas Young's double slit experiment

• Physics345
In summary, when a light is shone through a double slit, a pattern of bright and dark fringes is visible on the screen due to the overlapping of two waves that pass through the two barriers. Constructive interference occurs when peaks line up over peaks, creating bright fringes, while destructive interference occurs when peaks match up with valleys, creating dark fringes.Using Thomas Young's double-slit experiment, the wavelength of light being used to create the interference pattern can be determined in three different ways. The first method involves using the formula mλ=d sin⁡θ, where m is the order of the fringe, λ is the wavelength, d is the distance between the slits, and θ is the angle to the desired fringe
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.

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.

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

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

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.

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 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.

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.

Last edited:
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.

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.

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

## What is the purpose of Thomas Young's double slit experiment?

The purpose of Thomas Young's double slit experiment was to demonstrate the wave-like nature of light. It was also used to measure the wavelength of light, which was a major contribution to the understanding of light as a wave.

## How did Thomas Young's double slit experiment work?

In the experiment, Young shone a beam of light through a barrier with two slits, creating an interference pattern on a screen behind the barrier. This pattern showed that light behaves like a wave, as it exhibited areas of both constructive and destructive interference.

## What were the major findings of the experiment?

The major findings of the experiment were that light exhibits properties of both a particle and a wave. This was a significant discovery at the time, as it provided evidence for the wave theory of light and challenged the previous understanding of light as a purely particle-based phenomenon.

## What were the implications of Thomas Young's experiment on the understanding of light?

The experiment had a major impact on the understanding of light, as it showed that light has wave-like properties and can exhibit interference. This led to the development of the wave theory of light and furthered the understanding of the electromagnetic spectrum.

## How has Thomas Young's double slit experiment influenced modern science?

Thomas Young's double slit experiment has had a lasting impact on modern science, particularly in the field of quantum mechanics. It has been used to study the wave-particle duality of light, and has also been replicated with other particles such as electrons, further solidifying the concept. The experiment also paved the way for new discoveries and advancements in the understanding of light and its properties.

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