Interference, optical path, phase difference and path difference, interference due to reflected and transmitted light wave_TRUE NOTE

INTERFERENCE :

The non-uniform distribution of energy due to the super position of two light waves in a medium is known as the interference. OR, when the light waves from two different sources undergo superposition, there will be non-uniform distribution of the energy in the medium. This phenomenon of light is called interference. For the interference, two sources of light should be identical.

In the interference, there will be no loss of energy but energy is transformed from minimum displacement point to the maximum displacement point. The minimum and maximum displacement point appear alternatively.

Optical Path:

It is defined as the distance travel by the light in a vacuum for the given interval of time in medium. Let d be the distance travel by the light in a medium, then

Geometric path (d) = velocity (v) x time (t)

At the same time (t), the distance travelled by light in vacuum is called optical path.

i.e. Optical path (L) = c x t = c x \(\frac{d}{v} = \mu\) x d

therefore, Optical path (L) = Refractive index ( \(\mu\) ) x geometrical path (d)

RELATIONSHIP BETWEEN PHASE DIFFERENCE AND PATH DIFFERENCE:

We know that the waves travel through the distance of \(\lambda\), it makes phase angle of 2\(\pi\).

Let us suppose, for the small distance x, the phase angle is \(\delta\).

Then, \(\lambda\) \(\rightarrow\) 2\(\pi\)

1 \(\rightarrow\) \(\frac{2\pi}{\lambda}\)

X \(\rightarrow\) \(\frac{2\pi}{\lambda}\)x

Therefore, for the phase difference \(\delta\), path difference will be \(\frac{2\pi}{\lambda}\) x

i.e. \(\delta\) = \(\frac{2\pi}{\lambda}\)x

so, phase difference = \(\frac{2\pi}{\lambda}\) path difference

This is the required relation between phase difference and path difference.

Again, path difference = \(\frac{\lambda}{2\pi}\) path difference

1. For the bright point;

Phase difference = \(\frac{\lambda}{2\pi} 2n\pi\) = n\(\lambda\)

Therefore, path difference = n\(\lambda\)

2. For the dark point;

Path difference = \(\frac{\lambda}{2\pi}\) (2n+1)\(\pi\)

$$=\;\;\;(2n+1)\frac{\lambda}{2}$$

$$\therefore Path\; difference=(2n+1)\frac{\lambda}{2}$$

Interference Due To Reflection Of Light Wave :

The schematic diagram for the interference due to reflected light wave is show in the diagram below.

Consider a thin transparent film of refractive index \(\mu\) and thickness t. A light wave is allowed to incident on the upper surface at point B and certain parts of the incident beam gets reflection and remaining part is refracted. The refracted part is again refracted from the point C and it is finally transmitted along the direction of R. Here, we have to find the path difference between reflected light waves BQ and DR. Two perpendicular are drawn from the point B and D at points E and F as sown in the figure. The optical path difference between the reflected light waves is;

$$=\mu(BC+CD)-BF\dotsm(1)$$

We know, \(\mu\) = \(\frac{sin i}{sin r}\) = \(\frac{BF}{BD}X\frac{BD}{DE}\)

$$BF=\mu DE\dotsm(2)$$

Solving (1) and (2), we get;

Optical path difference = \(\mu\)[ BC + CD – DE ]

$$=\mu [ BC + CD - DE ]$$

[\(\therefore\) BC = PC ]

So, optical path difference = \(\mu\). PE ; Since PC + CD - ED = PE

Again, from the triangle BEP,

Cos r = \(\frac{PE}{BP}\)

Or, PE = BP Cos r

Therefore, PE = 2t Cos r [since, BP = 2t]

So, optical path difference = 2\(\mu t\)Cos r\(\dotsm\)(3)

The equation (3) can not give the correct path difference because one of the interfering beam has been reflected from denser medium to rarer medium. We know, when, when light reflects from the denser surface to rarer medium, then there exists an extra path difference of \(\frac{\lambda}{2}\) or equivalent phase difference of \(\pi\).

Therefore, the correct path difference due to reflected light wave is 2\(\mu\)t cosr - \(\frac{x}{2}\)

Interference Due To Transmitted Light Wave :

The diagram for the interference due to transmitted light wave is shown in the diagram below.

Consider a thin transparent film with refractive index \(\mu\) and thickness t. The light wave incident on the upper surface get reflected along BC and certain part of it is reflected along B whereas remaining part transmitted along the direction of CT. The light wave, the direction of CD is reflected from the point D and finally it is transmitted along E\(T_1\). Here we have to find the path difference between two transmitted light waves CT and D\(T_1\). The optical path difference between the reflected light waves is

$$=(DE + CD )\mu-CP$$

Also, we know, $$\mu=\frac{sini}{sinr}

$$=\frac{CP}{CE} \frac{CE}{QE}$$

Therefore, CP = \(\mu\) QE\(\dotsm\)(2)

Solving equation (1) and (2), we get,

Optical path difference = \(\mu\)[ CD + DE - MD ]

$$=\mu RQ$$

[\(\therefore\) CD + DE – MD = PM ]

Now, from the triangle RCQ,

Cosr = \(\frac{RQ}{RC}\)

Or, RQ = RC cos r

Or, RQ = 2t cos r

Therefore, optical path difference = 2\(\mu\)tcos r

So, condition of bright image is 2\(\mu\)tcos r = \(\lambda\)

And condition of dark image is 2\(\mu\)tcos r = (2n-1)\(\frac{\lambda}{2}\)

References:

Adhikari, P.B, Daya Nidhi Chhatkuli and Iswar Prasad Koirala. A Textbook of Physics. Vol. II. Kathmandu: Sukunda Pustak Bhawan, 2012.

Jenkins, F.A and H.E White. Fundamental of optics. New York (USA): McGraw-Hill Book Co, 1976.

wood, R.W. Physical Optics. New York (USA): Dover Publication , 1934.