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Optical Path Application in Young’s Double Slit Experiment

When light travels through a medium of thickness t, the optical path traveled is \mu t. When one of the slits in  YDSE is covered by a sheet, the air is replaced by the sheet. There fore, optical path changes by \left ( \mu -1 \right )t.

You have seen so many this type of article on Google but this will satisfied you from your inner soul.
I am going to discuss theory and some problems based on the concept of the optical path.

I hope you like it.

Optical Path Diagram

Suppose a parallel beam of light travelling in a vacuum is incident on surface AC of a medium of the refracting index \mu.\\AB is perpendicular to the incident rays and hence represents a wavefront of the incident light. Similarly, the CD is perpendicular to the refracted rays and represents a wavefront of refracted light. Again, the phase of the wave has constant value at different  points of a wavefront

so, the phase at A= Phase at B  &

Phase at C= Phase at D

so, the phase difference between A and B equals to phase difference between B and C

Using Snell’s law,

\; \; \; \frac{sin i}{sin r}=\mu\\

 

\Rightarrow \frac{\left ( \frac{BC}{AC} \right )}{\left ( \frac{AD}{AC} \right )}=\mu\\

 

\Rightarrow \frac{BC}{AD}=\mu\\

 

\Rightarrow BC=\mu \left ( AD \right )\\

 

The phase of lightwave changes by an equal amount whether it covers a distance BC=\mu \left ( AD \right ) in vacuum or AD in a medium.

Therefore, a path AD in a medium of refraction index \mu is equivalent to a path \mu \left ( AD \right ) in Vaccum which we call optical path.

The related problem on this topic is given below:

Q.(1)  A double-slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1mm and a distance between the plane of slits and screen is 1.33m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300A^{\circ}.

(a) Calculate the fringe width.

(b) One of the slite of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minimum on the axis.

Ans. (a)

fringe width =\beta=\frac{\lambda _{med}D}{d}\\

\therefore \beta = \frac{\lambda _{v}}{\mu _{m}}\cdot \frac{D}{d}= \frac{6300\times 10^{-10}}{1.33}\times \frac{1.33}{10^{-3}}\\

\beta = 0.63mm\\

Optical change in path

change in path difference if one of the slits of apparatus is covered by a thin glass sheet

\Delta x= \left ( \mu _{g}-\mu _{m} \right )t

for one wavelength  path differance, no. of  fringe shifted equals to one.

so, for path differance, \left ( \mu _{g}-\mu _{m} \right )t, no. of fringe shifted

N= \frac{\left ( \mu _{g}-\mu _{m} \right )t }{\lambda}\\

\therefore displacement of fringes on the screen = \frac{\left ( \mu _{g}-\mu _{m} \right )t }{\lambda}\beta .

According to our question:

\frac{\beta }{2}= \frac{\left ( \mu _{g}-\mu _{m} \right )t }{\lambda}\beta \\

\Rightarrow t= \frac{\lambda }{2\left ( \mu _{g}-\mu _{m} \right )}\\

\Rightarrow t=1.575\mu m. Ans.

Q (2) In young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate having the same thickness as the first one but having refractive index 1.7. The interference pattern is observed using the light of wavelength 5400A^{\circ}. It is found that point p on the screen where the central maximum(n=0) fell before the glass plates were inserted now has \frac{3}{4} the original intensity. It is further observed that what used to the fifth maximum earlier, lies below point P, while the sixth minimum lies above P. Calculate the thickness of the glass plate (Absorption of light by glass plate may be neglected).

Ans.

Optical Young's Slit Experiment

path difference between two waves starting from s_{1} and s_{2} reaching at point P,

\Delta x= \left ( \mu _{2}-\mu _{1} \right )t.

In the absence of plates, resulting  intensity at point P

\left ( I_{Resultant} \right )_{p}= 4I.

In the presence of plates,

\left [ \left ( I_{Resultant} \right )_{p} \right ]_{finally}= \frac{3}{4}\left [ \left ( I_{Resultant} \right )_{p} \right ]_{initially}

I+I+2\sqrt{I}\sqrt{I}cos\phi = \frac{3}{4}\times 4I\\

2I+2Icos\phi = 3I\\ 2I\left ( 1+cos\phi \right )= 3I\\ cos\phi = \frac{1}{2}\\ \Rightarrow cos\phi = cos\frac{\pi }{3}\\ \phi = 2m\pi \pm \frac{\pi }{3}\\

According to our question fifth maximum earlier lies below at point P, while the sixth minimum lies above point P

 so,\; \; \; \frac{11\lambda }{2}> \left ( \mu _{2}-\mu _{1} \right )t> 5\lambda \\

\frac{11\lambda }{2\left ( \mu _{2}-\mu _{1} \right )}> t> \frac{5\lambda }{\left ( \mu _{2}-\mu _{1} \right )}\\

\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \frac{11\lambda }{2\times 0.3}> t> \frac{5\lambda}{0.3}              …..(1)

Again, as we  know,

path difference= \frac{\lambda }{2\pi }\times phase difference.

\Rightarrow \left ( \mu _{2} -\mu _{1}\right )t= \frac{\lambda }{2\pi }\times \left ( 2m\pi \pm \frac{\pi }{3} \right )\\

0.3\: t= \frac{\lambda }{2\pi }\left ( 2m\pi \pm \frac{\pi }{3} \right ).\\

according to our question : m sould be equal to 5

\therefore 0.3t= \frac{\lambda }{2\pi }\left ( 2\times 5\pi \pm \frac{\pi }{3} \right )

Considering positive sign we get,

t= 93\times 10^{-7}m

Considering negative sign we get,

t= 87\times 10^{-7}m

but , from (1) : t= 9.3\mu m Ans.

 

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