Physics 2 Lecture 25

Wave superposition

• The resultant superposition of any two waves in space is given by the sum of the two constituent waves.
• This can leave to various interesting effects

- Example: Harmonic Traveling Waves

$$y(t)=y_m\sin (kx-\omega t)=y_m\sin (k[x-vt])$$

• This represents a wave of amplitude $y_m$ traveling in the x direction at wave speed $v_{phase}=\frac{\lambda }{t}=\lambda f$.
• The phase of the wave is given by the argument of the sin function. For the example above:

$$\phi (t)=kx-\omega t+\varphi$$

• Where $\varphi$ is the initial phase of the wave. A point with a specific phase moves to the right w/ velocity $v$.

Phase difference

• Phase difference, $\Delta \phi$, refers to how the argument of the sin function differs at different times or places in the wave.
• 2 times at the same point:

$$\Delta \phi =\left[\frac{2\pi }{\lambda }v\Delta t\right]=\omega \Delta t$$

• 2 points at the same time

$$\Delta \phi =\left[\frac{2\pi }{\lambda }\Delta x\right]=k\Delta x$$

Separating time and space

• Consider a plane wave:

$$E(x,t)=E_0\sin [\omega t-(kx+\varphi )]$$

• To separate the time and space elements, let $a(x,\varphi )=-kx-\varphi$
• In the case of two waves coexisting in space, we get:

$$E_1=E_{01}\sin (\omega t+{\alpha }_1)=E_{01}[\sin (\omega t)\cos {\alpha }_1+\cos (\omega t)\sin {\alpha }_1]$$

and

$$E_2=E_{02}\sin (\omega t+{\alpha }_2)=E_{02}[\sin (\omega t)\cos {\alpha }_2+\cos (\omega t)\sin {\alpha }_2]$$

And the traveling wave is given by

$$E_T=E_1+E_2=[E_{01}\cos {\alpha }_1+E_{02}\cos {\alpha }_2]\sin (\omega t)+[E_{01}\sin {\alpha }_1+E_{02}\sin {\alpha }_2]\cos (\omega t)$$

Adding and squaring the two terms gives:

$$E_0^2=E_{01}^2+E_{02}^2+2E_{01}{E}_{02}\cos ({\alpha }_2-{\alpha }_1)$$

Where the $2E_{01}{E}_{02}\cos ({\alpha }_2-{\alpha }_1)$ term is known as the interference term. Once we have the field, we can find the intensity:

$$I=\frac{E^2}{2{\mu }_{0}c}⟹I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \delta$$

where $\delta ={\alpha }_2-{\alpha }_1$

• The intensity of resulting wave is not just sum, but there is also interference term.

• There is constructive and destructive interference

• Special case $I_1=I_2$:

$$I=4I_1{\cos }^2\left(\frac{\delta }{2}\right)$$

Complex Expression of waves:

$$E_n=E_{0n}{e}^{j(\omega t+{\alpha }_1)}$$

Then

$$E_T=e^{j\omega t}[E_{01}{e}^{j{\alpha }_1}+E_{02}{e}^{j{\alpha }_2}]=E_0{e}^{jx}{e}^{j\omega t}$$

and intensity is given by:

$$I=(E_0{e}^{j\alpha })(E_0{e}^{j\alpha }{)}^{\ast }$$

Where

$$\tan \alpha =\frac{E_{01}\sin {\alpha }_1+E_{02}\sin {\alpha }_2}{E_{01}\cos {\alpha }_1+E_{02}\cos {\alpha }_2}$$

Cool thing about complex exponentials is that they can be graphically represented with Phasors. We can represent the superposition of many waves graphically as an addition of vectors in the complex plane. For cosine waves, $E_0$ will be the projection onto the horizontal axis and sine waves will be onto the vertical axis.

Since $a=-kx-ϵ$ where $k=\frac{2\pi }{\lambda }$ we can write $\delta ={\alpha }_2-{\alpha }_1=(k_1{x}_1+{ϵ}_1)-(k_2{x}_2+{ϵ}_2)$

$$=\left[\frac{2\pi }{{\lambda }_1}{x}_1-\frac{2\pi }{{\lambda }_2}{x}_2\right]+({ϵ}_1-{ϵ}_2)$$

What causes Phase Differences?

• Difference in wave generation
• Path length differences
• Reflections

To find total phase shift:

$$\delta ={\delta }_{initalphase}+{\delta }_{pathlength}+{\delta }_{reflection}$$ $$=ϵ+{\delta }_{opl}+{\delta }_R$$

Supposing the two waves are initially in phase $({ϵ}_1={ϵ}_2)$ we have:

$$\delta =\frac{2\pi }{{\lambda }_1}{k}_1-\frac{2\pi }{{\lambda }_2}{k}_2=\frac{2\pi }{{\Lambda }_0}{n}_1{L}_1-\frac{2\pi }{{\lambda }_0}{n}_2{L}_2$$ $$\delta =\frac{2\pi }{{\lambda }_0}[n_1{L}_1-n_2{L}_2]$$

Where the $[n_1{L}_1-n_2{L}_2]$ term is the optical path differences

Young’s Double Slit Experiment

Above is a diagram of Young’s Double Slit Experiment. R is distance from slit to screen, d is distance between slits. R >> d. Path length difference is given by $\Delta L=d\sin \theta$

• Constructive interference occurs when $\Delta L=d\sin \theta =m\lambda$
• Destructive Interference occurs when $\Delta L=d\sin \theta =\left(m+\frac{1}{2}\right)\lambda$ Where $m\in$ integers. It is known as the order of the maxima/minima. We can also use the small angle approximation for this:

$$\sin \theta \approx \tan \theta +\frac{y}{s}$$

in this limit, the position of the m’th bright fringe is

$$y_m=m\lambda \frac{D}{d}$$

or

$${\theta }_m=\frac{m\lambda }{d}$$

Then the distance between adjacent maxima is given by

$$\Delta y=y_{m+1}-y_m=\frac{(m+1)\lambda D}{d}-\frac{m\lambda D}{d}$$ $$⟹\Delta y=\frac{\lambda D}{d}$$

The intensity for the two slit interference is given by

$$I=4I_0{\cos }^2\left(\frac{\pi d}{\lambda D}y\right)$$