# Intensity and Interference Patterns (double slit)

• garthenar
In summary: If you leave out the factor of 2 in the formula, then I_0 would be the intensity of the central max due to one slit alone.
garthenar
Homework Statement
Two slits spaced 0.0720 mm apart are 0.800 m from a screen. Coherent light of wavelength λ passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is 3.00 mm. The intensity at the peak of the central maximum is 0.0900 W/m2.

a) What is the intensity at point on the screen that is 2.00 mm from the center of the central maximum?

b) What is the intensity at point on the screen that is 1.50 mm from the center of the central maximum?
Relevant Equations
λ=(ax)/D
a = distance between the two slits
x = the distance between maximums
m = which maximum your looking at (from the center)
D = the distance between the "source" (slits) and the screen

path difference
Δp = asin(θ)
= atan(θ)
= (ay/D)
y = vertical distance

phase difference
Φ = Δp (2π / λ)
= Δp k

I = 4(I_0)cos^2(Φ/2)
I'm still on part a.
I think that i may have the wrong equation for intensity.
I'm not sure I'm using the right numbers for the "first minimum".

I started with getting the wavelength

λ=(ax)/D
since the first minimum occurs at m = 0.5 I multiplied the distance to the first minimum by 2 to get the distance to the first "fringe". (point of maximum constructive interference)

I then got the path difference

Δp = (ay/D)
I used the distance to the point I want the intensity at for y

Then the phase difference

Φ = Δp (2π / λ)

And finally my intensity

I = 4(I_0)cos^2(Φ/2)

Which was wrong.
Can you help me figure out where I went wrong. I've tried several variation on this.

Thank you

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Your work looks good up to here:
garthenar said:
And finally my intensity

I = 4(I_0)cos^2(Φ/2)
View attachment 261531
Be sure you understand the meaning of I_0 in the formula. Is it the intensity of the central max of the double-slit pattern, or is it the intensity that one slit alone would produce at y = 0 on the screen?

TSny said:
Your work looks good up to here:
Be sure you understand the meaning of I_0 in the formula. Is it the intensity of the central max of the double-slit pattern, or is it the intensity that one slit alone would produce at y = 0 on the screen?
Wait, I though the central max was at y =0. Where is the central max?

garthenar said:
Wait, I though the central max was at y =0. Where is the central max?
Yes, the central max is at y = 0. The question is, what does the symbol I_0 represent in the formula?

TSny said:
Yes, the central max is at y = 0. The question is, what does the symbol I_0 represent in the formula?
I'm not entirely sure. My professor said that I should use the equation (I_0)cos^2(Φ/2) with I_0 equaling the intensity at the origin of the two slit model. So I'm assuming that the 4(I_0) has something to do with the the intensity one slit would produce. We haven't done single slits so I'll have to go check that out.

garthenar said:
I'm not entirely sure. My professor said that I should use the equation (I_0)cos^2(Φ/2) with I_0 equaling the intensity at the origin of the two slit model. So I'm assuming that the 4(I_0) has something to do with the the intensity one slit would produce. We haven't done single slits so I'll have to go check that out.
Yes. In the 4(I_0) formula, I_0 is the intensity that one slit alone would produce at the location of the central max. The total amplitude at the central maximum due to both slits is twice the amplitude due to one slit alone. Intensity is proportional to the square of the amplitude. So, the intensity of the central max due to both slits together is 4 times the intensity that one slit alone would produce.

If you leave out the factor of 4 in the formula, then I_0 would be the intensity of the central max due to both slits together.

## 1. What is the double slit experiment?

The double slit experiment is a famous experiment in physics that demonstrates the wave-like nature of light. It involves shining a beam of light through two parallel slits and observing the resulting interference pattern on a screen.

## 2. What is the relationship between intensity and interference patterns?

The intensity of light refers to the amount of energy that passes through a given area in a given amount of time. In the double slit experiment, the intensity of the light affects the visibility and spacing of the interference pattern. Higher intensity light will result in a brighter and more defined interference pattern, while lower intensity light will result in a dimmer and less defined pattern.

## 3. What causes interference patterns to form in the double slit experiment?

The interference patterns in the double slit experiment are caused by the superposition of two or more waves. When the waves pass through the two slits, they interfere with each other, resulting in areas of constructive interference (where the waves add up and create a brighter spot) and areas of destructive interference (where the waves cancel each other out and create a darker spot).

## 4. How does the distance between the slits affect the interference pattern?

The distance between the slits plays a crucial role in the formation of the interference pattern. As the distance between the slits increases, the spacing of the bright and dark fringes in the interference pattern also increases. This is because the waves coming from the two slits have to travel a greater distance before they reach the screen, resulting in a larger difference in their phase.

## 5. Can the double slit experiment be used to study other types of waves besides light?

Yes, the double slit experiment can be used to study other types of waves, such as sound waves and water waves. In fact, the experiment was first performed with water waves by Thomas Young in 1803. It has since been replicated with other types of waves, further supporting the wave-like nature of light.

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