Optics - index of refraction and phase lag

[atomqwerty]

Homework Statement

A light beam of wavelenght 500nm travels from A to B in vacuum. (AB = 10mm)
If we put a glass sheet with n=1.5 and 1mm thick, calculate how change the number of periods. Calculate the phase lag too.

Homework Equations

k0 = 2Pi/λ (vacuum)

k= k0·n

The Attempt at a Solution

If in the vacuum the wave gas a wavelenght λ=2Pi/k0, into a n=1.5 glass, it will have λ'=2Pi·n/k0 (?)

Thanks

 

[zhermes]

Yes, that's correct.

 

[SammyS]

How many more waves are there in 1 mm of glass than there are in 1 mm of free space?

 

[atomqwerty]

How many more waves are there in 1 mm of glass than there are in 1 mm of free space?

If the wavelenght is λ, in a free space of d=1mm there will be k=λd waves, and in d'=1mm glass (index n) there will be k'=λnd', so if we put the glass into de space d, this leads to

k=λ(d-d')

k'=λd'n

therefore the number of waves (total) will be

K = k+k'

Is this correct?

thanks

 

[vela]

No, you didn't calculate the number of cycles correctly. Both λ and d have units of length, so using your expression, k would have units of length squared. The number, however, should be unitless.

(k isn't the best letter to use since k is typically used to denote the spatial frequency.)

 

[atomqwerty]

No, you didn't calculate the number of cycles correctly. Both λ and d have units of length, so using your expression, k would have units of length squared. The number, however, should be unitless.

(k isn't the best letter to use since k is typically used to denote the spatial frequency.)

If k = 2Pi/λ [meters^(-1)], then kd, with [d=meters] is unitless, isn't it?

Thanks

 

[atomqwerty]

(k isn't the best letter to use since k is typically used to denote the spatial frequency.)

Ok, my mistake, but is K = k+k' as we define k and k' [uniteless] the number of waves?

 

[SammyS]

How many more waves are there in 1 mm of glass than there are in 1 mm of free space?

The number of waves in a thickness d of glass, is: d/(λ) = d/(λ₀/n) = nd/λ₀, where λ₀ is the wavelength in a vacuum.

So, the number of EXTRA cycles is: nd/λ₀-d/λ₀ = (n-1)(d/λ₀)