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I know that one of the causes of a phase difference in waves is due to a path difference. My textbook states that a phase difference could be due to waves travelling in different mediums and that when that happens we have to keep track of the cycles in each meduim.

I don't understand how we can keep track of the cycles, is that even possible? Can someone please elaborate further from the statement given in my textbook so that i can get a bit of insight. Thank you.

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  • $\begingroup$ An interferometer measures phase difference between two waves. Phase difference is usually only meaningful when comparing two waves that have exactly the same frequency. $\endgroup$
    – S. McGrew
    Aug 19, 2018 at 21:19

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See how you get on with this worked example…

A glass plate of thickness 0.0100 mm is placed in the path of a laser beam. The beam hits the plate near its edge, so that part of the beam travels through the plate and the rest misses the plate and travels through air. We will ignore diffraction effects. Suppose that the wavelength of the laser in air is exactly 633.0 nm, the refractive index of air is 1.000 and the refractive index of glass is 1.520.

The number of wavelengths of the laser light in 0.0100 mm of air is $\frac{0.0100\ \text {mm}}{633.0\ \text{nm}}=15.8$.

But the wavelength of the light in the glass will be $\frac{633.0\ \text{nm}}{1.520}=416.4\ \text{nm}$ because light travels more slowly in the glass by a factor of 1.520. Therefore the number of wavelengths of the laser light in 0.10 mm of glass is $\frac{0.0100\ \text {mm}}{416.4\ \text{nm}}=24.0$.

But travelling in glass doesn't change the frequency of the light (you can't lose or gain cycles) nor does it change the periodic time. So the light that has travelled through the air will be 8.2 cycles ahead in time of the light that has travelled through the glass.

Alternative method

time for light to travel 0.0100 mm through air = $\frac{0.0100\ \text{mm}}{c}.$

time for light to travel 0.0100 mm through glass (in which it travels more slowly by a factor of 1.520) = $\frac{1.520 \times 0.0100\ \text{mm}}{c}.$

Therefore extra time taken through glass = $\frac{0.520 \times 0.0100\ \text{mm}}{c}.$

But time per cycle = $\frac{1}{f}=\frac{\lambda_{air}}{c}=\frac{633\ \text{nm}}{c}$.

So number of cycles lag by going through glass is$$\frac{0.520 \times 0.0100\ \text{mm}}{c}\ \div\ \frac{633\ \text{nm}}{c}=8.2$$

If you prefer, you can substitute the numerical value for $c$, the speed of light, and work out the times as you go along. I didn't do this, as I wanted to show you that $c$ cancels, that is you don't need to know its value.

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  • $\begingroup$ I am not getting the part of frequency lag. If no cycles are gained or lost as you stated, why then does the light that travels through glass lag by a number of cycles. Isn't that like losing cycles. I am confused. $\endgroup$
    – Energy
    Aug 20, 2018 at 2:57
  • $\begingroup$ And is it ok to compare the number of wavelengths of these waves. I am asking because one wave experiances huge wavelengths while the other experiances short wavelenths for a given period of time. Thank you. $\endgroup$
    – Energy
    Aug 20, 2018 at 3:12
  • $\begingroup$ Frequency is number of cycles $per\ second$. That's what's the same in glass as in air, not number of cycles. $\endgroup$ Aug 20, 2018 at 6:55
  • $\begingroup$ Neither wavelength is huge. The wavelength in air is, in my example, 1.520 times that in glass. Both are only hundreds of nanometers! $\endgroup$ Aug 20, 2018 at 6:57
  • $\begingroup$ Have you learnt the basics about how waves travel, things like definitions of wavelength, frequency, wave speed, $v=f \lambda$, refractive index (for light)? No chance of your understanding my answer unless you have. $\endgroup$ Aug 20, 2018 at 7:00

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