# Confusion regarding path difference by glass lab 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

## 1 Answer

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 $$t_{\rm glass}$$ assume there are $$N$$ waves each of wavelength $$\lambda_{\rm glass},\, \Rightarrow t_{\rm glass} = N\,\lambda_{\rm glass}$$.

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

The difference in thickness containing the same number of waves is $$t_{\rm medium}- t_{\rm glass} = N\,\lambda_{\rm medium} - N\,\lambda_{\rm glass}$$.

$$f \lambda_{\rm medium} = c_{\rm medium} = \dfrac {c_{\rm vacuum}}{n_{\rm medium}}$$ and $$f \lambda_{\rm glass} = c_{\rm glass} = \dfrac {c_{\rm glass}}{n_{\rm glass}}$$ where $$f$$ is the frequency, $$c$$ is the wave speed and $$n$$ is the refractive index $$\Rightarrow \lambda_{\rm glass} n_{\rm glass} = \lambda_{\rm medium} n_{\rm medium}$$

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

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

If the number of extra waves had been $$\frac 12$$ 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 $$\lambda_{\rm medium} = \dfrac {\lambda_{\rm vacuum}}{n_{\rm medium}}$$ into the equation the number of extra waves is $$\dfrac{(n_{\rm glass} - n_{\rm medium})t_{\rm glass}}{\lambda_{\rm vacuum}}$$

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

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

So in summery:

$$\left (\dfrac{n_{\rm glass}}{n_{\rm medium }}-1\right )t_{\rm glass}$$ relates to the wavelength in the medium

and

$$(n_{\rm glass}-n_{\rm medium })\,t_{\rm glass}$$ relates to the wavelength in a vacuum.