Phase shift from path difference

In summary, to determine the lengths at which constructive and destructive interference occur in a glass block with different refractive indices, one must calculate the phase shift of 2 rays passing through the block and set it equal to an integer or half-integer multiple of 2pi. This results in the conclusion that constructive interference occurs at a length of (7/2.5*10^6) meters, while destructive interference occurs at a length of (3.5/2.5*10^6) meters. The different refractive indices in the upper and lower halves of the glass block are the cause of this phase difference.
  • #1
kankerfist
35
0
1. Homework Statement

A wave front moving horizontally encounters a block of glass that has an index of refraction of 1.50 in its upper half and an index of refraction of 1.25 in its lower half. The wavelength of the wave front in air is 700 nm. At what lengths would the glass block cause constructive interference (bright spots) and destructive interference (dark spots)?


2. Homework Equations

n1=1.50
n2=1.25
λair=700E-9 m


3. The Attempt at a Solution

I came up with the following when I encountered this study problem and I was hoping someone could double check to see if I did this correctly:

Constructive interference should occur when the glass block length causes a phase shift of (integer)(2pi). When 2 rays leave the block they will be phase shifted by (2pi)(length of block)/(λ in n1 and n2). So the total phase shift for 2 rays going through the block:

(2pi)(L)/(λ in n1) - (2pi)(L)/(λ in n2)

and:
λ in n1= λ in air / n1 = (700E-9)/(1.5)
λ in n2 = (700E-9)/(1.25)

so constructive interference should occur when:

(2pi)(L) / [(700E-9)/(1.5)] -
(2pi)(L) / [(700E-9)/(1.25)] =
(integer)(2pi)

this reduces to:

Length of glass = 7(integer)/2.5E+6

So constructive interference should occur when the length of the glass block is any positive integer times (7/2.5*10^6) in meters. If anybody could take a glance and see if I messed up somewhere i'd appreciate it!
 
Physics news on Phys.org
  • #2


I can confirm that your approach and calculations are correct. Constructive interference will occur when the length of the glass block is a positive integer times (7/2.5*10^6) meters, resulting in a phase shift of (integer)(2pi). Similarly, destructive interference will occur when the length of the glass block is a half-integer times (7/2.5*10^6) meters, resulting in a phase shift of (half-integer)(2pi). This is due to the different refractive indices in the upper and lower halves of the glass block, which cause a phase difference between the two rays passing through the block. Keep up the good work!
 

Related to Phase shift from path difference

1. What is a phase shift from path difference?

A phase shift from path difference is a phenomenon that occurs when two waves with slightly different frequencies interfere with each other. This results in a change in the relative phase or timing of the waves, causing a shift in the overall waveform.

2. How does the path difference affect the phase shift?

The path difference between two waves refers to the difference in distance traveled by each wave to reach a certain point. This difference in distance results in a difference in phase, which leads to a phase shift when the waves interfere with each other.

3. What is the relationship between phase shift and wavelength?

The phase shift is directly proportional to the wavelength of the waves. This means that as the wavelength increases, the phase shift also increases. Conversely, as the wavelength decreases, the phase shift decreases as well.

4. How is the phase shift from path difference calculated?

The phase shift from path difference can be calculated using the formula Δϕ = 2πΔx/λ, where Δϕ is the phase shift, Δx is the path difference, and λ is the wavelength of the waves.

5. What are some real-world applications of phase shift from path difference?

The phase shift from path difference has a wide range of applications in various fields, including telecommunications, audio and radio signal processing, and medical imaging. It is also essential in understanding and studying wave phenomena, such as diffraction and interference.

Similar threads

  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Quantum Physics
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
3K
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
934
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
8K
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Back
Top