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

AI Thread 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 by the refractive index. The wavelength in the medium is found by dividing the free space wavelength by the refractive index. The phase difference is calculated using the formula dπ/λ, leading to a result of 8π/3, which raises questions about the reasoning behind the use of π and the nature of interference. Overall, the calculations emphasize the relationship between light behavior in different media and the impact of refractive index on phase.
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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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