Phase Diff. of 600 nm Light in Medium w/ Ref. Index 1.5

In summary, the given values are a wavelength of 600 nm and a refractive index of 1.5. The optical path length is 2.4 x 10^-6 m, the wavelength in the medium is 400 nm, and the phase difference after moving this distance is 8π/3. The phase difference was calculated using dπ/λ, with d being the path length and λ being the wavelength. This results in a phase difference of 8π/3, indicating destructive interference.
  • #1
charmedbeauty
271
0

Homework Statement



Light of free space wavelength 600 nm (6000 Å) travels 1.6 10
-6
m in a medium of
index of refraction 1.5.
Find:
1. the optical path length, [This is the physical path length x refractive index.]
2. the wavelength in the medium, and
3. the phase difference after moving that distance, with respect to light traveling the
same distance in free space.




Homework Equations





The Attempt at a Solution



just for part 3 I got the right answer but I am just a little confused if my resoning is right.

the phase difference I calculated using d∏/λ I used Pi since the refractive index was not an integer so I thought it would be destructive intefeerence.

The answer comes out to be 8∏/3 but I am a little confused on why?
 
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  • #2
Can you calculate the path length through that medium? (express your answer in wavelengths)
 
  • #3
NascentOxygen said:
Can you calculate the path length through that medium? (express your answer in wavelengths)

path length= actual length × (refractive index)

→ pathlength/wavelength.

= λ wavelengths


??
 

Related to Phase Diff. of 600 nm Light in Medium w/ Ref. Index 1.5

1. What is phase difference?

Phase difference is the measure of the difference in phase between two waves, typically measured in degrees or radians. It represents the relative position of two waves at a given point in time.

2. How is phase difference of 600 nm light calculated?

The phase difference of 600 nm light can be calculated using the formula Δφ = 2πΔnL/λ, where Δφ is the phase difference, Δn is the difference in refractive index between two mediums, L is the distance traveled by the light, and λ is the wavelength of the light.

3. What is the significance of a refractive index of 1.5 in this scenario?

The refractive index of 1.5 represents the optical density of the medium through which the light is passing. It affects the speed of light and the amount of phase difference that occurs as the light travels through the medium.

4. How does the phase difference of 600 nm light change in different mediums with varying refractive indexes?

The phase difference of 600 nm light will change in accordance with the refractive index of the medium it is passing through. As the refractive index increases, the phase difference will also increase.

5. How is phase difference of 600 nm light used in practical applications?

The phase difference of 600 nm light is important in a variety of applications, including optical sensors, interferometers, and telecommunications. It is also used in the study of wave behavior and can help determine the properties of different materials.

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