I came across the concept of optical path length a few days ago ; a case where parallel beams of light passing through a rectangular prism generate a path difference with respect to the rays not falling on the prism but going parallel sideways.
I know that the time lag between the two sets of rays cause the path difference …
1.But how do we apply the same to a triangular prism (how is the formula for path difference as stated in the above figure derived)..
2.Also what is actually optical path length
Guys please respond a bit early ..I have my exams coming in a few days
Best Answer
You need to count the waves.
Let the wavelengths be $\lambda_{\rm air}$ and $\lambda_{\rm glass}$
To have the same number of waves in $\rm AB$ in air as $\rm CD$ in glass the following equation must be true
$\dfrac {\rm AB}{\lambda_{\rm air}} = \dfrac {\rm CD}{\lambda_{\rm glass}} $
However $\lambda_{\rm glass} = \dfrac {\lambda_{\rm air}}{\mu_{\rm glass}} $ where $\mu_{\rm glass}$ is the refractive index of glass.
Putting this into the equation produces $\rm AB = \mu_{\rm glass} \,CD$
$\mu_{\rm glass} \,\rm CD$ is called the optical path length and contains the same number of waves as a length $\rm AB$ of air.
So your $\Delta x$ is the difference between:
the length of air which contains the same number of waves as a length of glass $\rm QS$
and
the length of air $\rm PR$.
If there two happened to be the same then if the waves left $P$ and $Q$ in phase then they must have arrived at $R$ and $S$ in phase.