Rays from the same beam passing through the prism

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

  • $\begingroup$ It will be highly appreciated if someone can just provide me up with the derivation of the path length written in the picture above.(only this particular case) $\endgroup$ – Piyush Kumar Apr 22 '17 at 12:29

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 $\rm QS$


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.

  1. You apply the rules of geometry to the situation. The optical path length PR must be the same as QS. As you have written it, $\Delta x = 0$.

  2. Optical path length is distance times refractive index. Another way of saying this is that it is the distance that the ray would travel in vacuum in the same time.

In a fixed time interval $t$ light travels a geometrical distance $x_1=vt$ in an optical medium, where $v=c/n$. Here $c$ is speed in vacuum and $n$ is the refractive index. $v$ is the speed in the medium, which is less than in vacuum.

In the same time interval light travels a distance $x_0=ct=c(x_1/v)=(c/v)x_1=nx_1$ in a vacuum.

So light takes the same time to travel a distance of $x_0=nx_1$ in vacuum as it takes to travel a distance $x_1$ in a medium of index $n$.

  • $\begingroup$ Okay if i apply a bit of geometry ...I do get it as zero but i want to ask how did we get that expression for path difference? $\endgroup$ – Piyush Kumar Apr 22 '17 at 12:15
  • $\begingroup$ The wavelength and velocity of the light rays are different in different media, but the frequency is always the same. (You cannot create or destroy waves except at source or sink. Like current in an electric circuit, the number of wave peaks passing a point every second must be the same all along the ray.) Path length is a way of comparing how many wavelengths there are in a given distance. So a distance of 10cm in glass contains 1.5 times as many wavelengths in glass than 10cm in vaccuum. This is because wave speed is lower in glass, so wavelength is also lower. $\endgroup$ – sammy gerbil Apr 22 '17 at 12:26
  • $\begingroup$ en.wikipedia.org/wiki/Optical_path_length $\endgroup$ – sammy gerbil Apr 22 '17 at 12:27
  • $\begingroup$ The link you provided has a direct formula for OPD .How is that derived? $\endgroup$ – Piyush Kumar Apr 22 '17 at 12:31
  • $\begingroup$ Thx a lot Sammy ..I get the derivation now but unfortunately i am still unable to write that expression i.e ∆x = n (QS) - (PR) in this particular figure $\endgroup$ – Piyush Kumar Apr 22 '17 at 13:30

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