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

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SUMMARY

The discussion focuses on calculating the optical path length, wavelength in a medium, and phase difference for light traveling through a medium with a refractive index of 1.5. The optical path length is determined by multiplying the physical path length (1.6 x 10-6 m) by the refractive index, resulting in a value that can be expressed in wavelengths. The wavelength in the medium is derived from the original wavelength of 600 nm, leading to a phase difference calculation of 8π/3 using the formula dπ/λ, where d is the distance traveled and λ is the wavelength in the medium.

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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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Can you calculate the path length through that medium? (express your answer in wavelengths)
 
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


??
 

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