# Confusion regarding path difference by glass slab in YDSE

If in a YDSE experiment the setup is immersed in a medium $$n_{med}$$, and a glass slab of thickness t of another medium $$n_{glass}$$ is introduced in front of one slit, then is the path difference due to introduction of the glass slab

$$(\frac{n_{glass}}{n_{med}}-1)t$$

or

$$(n_{glass}-n_{med})t$$

Attempt at solution: Since the optical path in the region of the glass slab, 'with the slab' is $$(\frac{n_{glass}}{n_{med}})t$$ and 'without the slab' is just t ; the path difference due to introduction of the glass slab should be

$$(\frac{n_{glass}}{n_{med}}-1)t$$

But

In a similar problem (given below) the path difference due to the slab alone according to my text is $$(n_{3}-n_{2})t$$

Why is this so? since both equations give different answers, What is the correct one?

NB: i know that similar questions have been asked but none have a proper complete answer hence i am asking this

I will assume that all the refractive indices that you have quoted in your question are relative to a vacuum.

The ideas of optical path and optical path difference is really all to do with counting wavelengths.

Within a thickness of glass $$t_{\text{glass}}$$, assume there are $$N$$ waves each of wavelength $$\lambda_{\text{glass}}$$. Therefore, we have $$t_{\text{glass}} = N\lambda_{\text{glass}}$$.

The thickness of medium $$t_{\text{medium}}$$ which has the same number of waves $$N$$, each of wavelength $$\lambda_{\text{medium}}$$, as before is $$t_{\text{medium}} = N\lambda_{\text{medium}}$$

The difference in thickness containing the same number of waves is $$t_{\text{medium}}- t_{\text{glass}} = N\lambda_{\text{glass}} - N\lambda_{\text{glass}}$$. Therefore, $$f \lambda_{\text{medium}} = c_{\text{medium}} = \frac {c_{\text{vacuum}}}{n_{\text{medium}}} \\ f \lambda_{\text{glass}} = c_{\text{glass}} = \frac {c_{\text{vacuum}}}{n_{\text{glass}}}$$ where $$f$$ is the frequency, $$c$$ is the wave speed and $$n$$ is the refractive index. Thus, $$\lambda_{\text{glass}} n_{\text{glass}} = \lambda_{\text{medium}} n_{\text{medium}}$$.

Substituting into the equation for difference in thickness gives $$t_{\text{medium}}- t_{\text{glass}}= \left(\frac{n_{\text{glass}}}{n_{\text{medium}}}-1 \right)t_{\text{glass}}$$

So the number of extra waves introduced by the insertion of the glass is $$\frac{t_{\text{medium}}- t_{\text{glass}} }{\lambda_{\text{medium}}}=\left (\frac{n_{\text{glass}}}{n_{\text{medium}}}-1 \right)\frac{t_{\text{glass}}}{\lambda_{\text{medium}}}$$

Had the number of extra waves been $$1/2$$, then the fringe pattern would have moved by half a fringe and where there had been a maximum there would be a minimum etc.

If you substitute $$\lambda_{\text{medium}} = \dfrac{\lambda_{\text{vacuum}}}{n_{\text{medium}}}$$ into the equation, the number of extra waves is $$\frac{(n_{\text{glass}} - n_{\text{medium}})t_{\text{glass}}}{\lambda_{\text{vacuum}}}$$

The term $$(n_{\text{glass}} - n_{\text{medium}})t_{\text{glass}}$$ is called the optical path difference and relates to the wavelength in a vacuum.

A thickness of glass $$t_{\text{glass}}$$ has the same number of waves as a thickness of vacuum equal to $$n_{\text{glass}} t_{\text{glass}}$$ and this is called the optical path length.

So in summary:

$$\left(\frac{n_{\text{glass}}}{n_{\text{medium}}}-1\right)t_{\text{glass}}$$ relates to the wavelength in the medium, and $$(n_{\text{glass}}-n_{\text{medium}})\,t_{\text{glass}}$$ relates to the wavelength in a vacuum.

• thanks a lot :D May 12 '19 at 11:25