Notes on the Phasor Diagram Method To determine the intensity of a 2-slits interfernce (or 2 -"point"source interference), one needs to add two sinusoidal waves with a relative phase.

Etot (t) = E1(t) + E2 (t) = E cos(ωt +φ) + E cos(ωt), where the relative phase is due to the path length difference.

According to your textbook, Etot (t) can be represented

by a single sinusoidal wave, i.e. Etot (t) = Ep cos(ωt +δ ). Your texbook proceeded immediately to show you how to determine

Ep using the "Phasor Diagram" method (as a recipe). What I would like to do:

(1) Derive (tediously) E cos(ωt +φ) + E cos(ωt) = Ep cos(ωt +δ )

using trig. identities and deduce Epand δ.

(2) Show you the easy way to find Epand δ using Phasor Diagram. (3) Show you the mathematical justification of the Phasor Diagram method using Euler's identity: eiθ = cosθ + isinθ Analysis of interference of 2 point sources.

At an arbitrary point P, the combined electric fields is that emited by source 1 and source 2: P r1 r1 r2 Ep (t) = E cos(ωt + 2π ) + E cos(ωt + 2π ) λ λ r (We have assumed that the two sources have the same amplitude, 2 which is not exactly true because one source is slightly further than the other source) Rewrite the expression to show explicitly the relative phase: r (r − r ) r E (t) = E cos(ωt + 2π 2 + 2π 1 2 ) + E cos(ωt + 2π 2 ) p λ λ λ r The common phase 2π 2 can be absorbed in starting time of your clock) λ

⇒ Ep (t) = E cos(ωt +φ) + E cos(ωt), where (r − r ) φ = 2π 1 2 is relative phase difference due to path length difference. λ Note: φ is in radian Tedious Method

Ep (t) = E cos(ωt +φ) + E cos(ωt) = [E cos(ωt)cosφ − E sin(ωt)sinφ]+ E cos(ωt) = [1+ cosφ]E cos(ωt) − E sinφ sin(ωt) ≡ Acos(ωt) − Bsin(ωt) A ≡ [1+ cosφ]E, B ≡ E sinφ

2 2 ⎡ A B ⎤ = A + B ⎢ cos(ωt) − sin(ωt)⎥ ⎣ A2 + B2 A2 + B2 ⎦ A B B sinφ Define : = cosδ, = sinδ ⇒ tanδ = = A2 + B2 A2 + B2 A [1+ cosφ] A2 + B2 = E [1+ cosφ]2 + sin2 φ = E 2(1+ cosφ)

2 ⎛ φ ⎞ ⎛ φ ⎞ = E 4 cos = 2E cos ⎝⎜ 2⎠⎟ ⎝⎜ 2⎠⎟ ⎛ φ ⎞ B sinφ ⇒ Ep (t) = 2E cos⎜ ⎟ cos(ωt +δ ); tanδ = = ⎝ 2⎠ A [1+ cosφ] Phasor Diagram –Easy “recipe”

Draw a vector with magnitude=E and makes an angle =ωt wrt the horizontal axis such that the horizontal component is Ecosωt.

Draw a vector with magnitude=E and makes an angle =ωt+φ wrt the horizontal axis such that the horizontal component is Ecos(ωt+φ).

Add these two vectors. The resultant vector

φ will have the correct amplitude = 2E|cos( 2 ) | sinφ and will have the correct angle =ωt+δ;tanδ = 1+ cosφ Equating the horizontal component of this vector to the horzontal components of the two oringial vectors provides the desired result:

φ Ecosωt+Ecos(ωt+φ)=2E|cos( 2 ) | cos(ωt +δ ) Use Euler’s Identy to explain the Phasor Diagram Method

Euler's identity: eiθ = cosθ + isinθ Hence, the complex number eiθ can be viewed as an unit vector in the complex plane, i.e. the horizontal component of the vector is cosθ, the vertical component of the vector is sin θ, and the magnitude of the vector is cos2θ + sin2 θ = 1. ⇒ Eeiωt = E[cosωt + isinωt] is the first vector on the phasor diagram Eei(ωt+φ ) = E[cos(ωt +φ) + isin(ωt +φ)] is the second vector. The resultant vector is the sum of these two vectors.

Q. What does the additon of these vectors, Eeiωt and Eei(ωt+φ ), have to do with the Ecosωt + E cos(ωt +φ)? See next slide. The real part of (Eeiωt + Eei(ωt+φ ) ) = E cosωt + E cos(ωt +φ) It turns out that it easier to add the two complex numbers first and then take the real part later.

The phasor diagram method helps you add the two complex numbers geometically as vectors.

But we can also add the two complex number algebraically: Eeiωt + Eei(ωt+φ ) = Eeiωt [1+ eiφ ] = Eeiωt [1+ cosφ + isinφ]

Any complex number A+iB can be written as its magnitude times an unit vector

2 2 2 2 iδ B A + B [cosδ + isinδ ] = A + B e ;tanδ = A

2 2 iδ iδ φ iδ ⇒ [1+ cosφ + isinφ] = [1+ cosφ] + sin φe = [2 + 2cosφ]e = 2 | cos 2 | e sinφ tanδ = 1+ cosφ

iωt i(ωt+φ ) φ iωt iδ ⇒ Ee + Ee = 2E | cos 2 | e e φ The real part is 2E | cos 2 | cos(ωt +δ )

It seems that adding these two complex number algebraically is as tedious as the original "tedious" method. But,when you have to add more cosine functions, then the complex number method (algbraically) or the phasor method (geometrically) will be easier.