Phase shift of light incident on 2 slits

In summary, at a point y = 2.8 mm from the central maximum, the intensity of the light is I = 55 W/m2.
  • #1
gsmtiger18
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0

Homework Statement


Light of wavelength λ = 450 nm is incident upon two thin slits that are separated by a distance d = 25 μm. The light hits a screen L = 2.5 m from the screen. It is observed that at a point y = 2.8 mm from the central maximum the intensity of the light is I = 55 W/m2.

a) Write an equation for the phase shift between the light from the two slits at the observation point in terms of the given variables.
ANSWER: Φ = 2πdy/(λL)

b) For the given data, what is the phase shift, in radians?
ANSWER: 0.3909 radians

c) What is the intensity of the light at the two slits Io in W/m2?

Homework Equations


Small angle approximation: y = mλ/d (for bright fringes)
Young's equation: Φ = kdsinθ
Path Difference Equation: ΔX = λΔΦ/2π

The Attempt at a Solution


I've worked out both of the first two parts, and the answers are correct. I'm not sure how to get the original intensity though. I tried dividing the intensity I at y by the phase shift Φ, which yielded 140.82 W/m2, but for some reason that's wrong.
 
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  • #2
There exists a formula giving the intensity of the interference pattern as a function of the phase shift. It goes like this (for 2 slits) :

I=4Iocos2(Φ/2)
 
  • #3
Benoit said:
There exists a formula giving the intensity of the interference pattern as a function of the phase shift. It goes like this (for 2 slits) :

I=4Iocos2(Φ/2)
Is there a way to derive that?
 
  • #4
Well, you know that light is an oscillating wave made of an electric and magnetic field. When you add two waves, like in Young's experiment, you could look at it as if you add up the electric fields of the waves. So add :

E1sint(wt+Φ1) and E1sin(wt+Φ2)

With a little work, you end up with :

E2=2E1cos(ΔΦ/2)sin(wt+ ((Φ12)/2)

So you know that the amplitude of this electric field is 2E1cos(ΔΦ/2).
We almost have it. You also know, maybe from your courses on mechanical waves, that power is proportional to the square of the amplitude. We have Io=k*E12

Therefore, I=k*E22=k*4E12cos2(ΔΦ/2)

Finally, because Io=k*E12, I=4Iocos2(ΔΦ/2)
 

What is the phase shift of light incident on 2 slits?

The phase shift of light incident on 2 slits refers to the difference in the phase of the waves passing through each of the two slits. This phase shift can affect the interference pattern that is created when the waves from the two slits overlap.

How does the distance between the two slits affect the phase shift?

The distance between the two slits can affect the phase shift by changing the path length difference between the two waves. This, in turn, can alter the interference pattern that is created.

What is the mathematical formula for calculating the phase shift?

The mathematical formula for calculating the phase shift is given by Φ = 2πd/λ, where Φ is the phase shift, d is the distance between the two slits, and λ is the wavelength of the incident light.

Can the phase shift be controlled or manipulated?

Yes, the phase shift can be controlled or manipulated by changing the distance between the two slits or by altering the wavelength of the incident light. This can be done using tools such as a diffraction grating or a prism.

What is the significance of the phase shift in experiments involving 2 slits?

The phase shift is significant in experiments involving 2 slits because it directly affects the resulting interference pattern. By understanding and controlling the phase shift, scientists can gain valuable insights into the behavior of light and its properties.

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