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(ω t + φ ) + E cos(ω t),
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) = E p cos(ω t + δ ).
Your texbook proceeded immediately to show you how to determine
E p using the "Phasor Diagram" method (as a recipe).
What I would like to do:
(1) Derive (tediously) E cos(ω t + φ ) + E cos(ω t) = E p cos(ω t + δ )
using trig. identities and deduce E p and δ .
(2) Show you the easy way to find E p and δ using Phasor Diagram.
(3) Show you the mathematical justification of the
Phasor Diagram method using Euler's identity: eiθ = cosθ + i sin θ
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:
r
r
E p (t) = E cos(ω t + 2π 1 ) + E cos(ω t + 2π 2 )
λ
λ
(We have assumed that the two sources have the same amplitude,
which is not exactly true because one source is slightly further than
the other source)
Rewrite the expression to show explicitly the relative phase:
r2
(r − r )
r
+ 2π 1 2 ) + E cos(ω t + 2π 2 )
λ
λ
λ
r
The common phase 2π 2 can be absorbed in starting time of your clock)
λ
⇒ E p (t) = E cos(ω t + φ ) + E cos(ω t), where
E p (t) = E cos(ω t + 2π
(r1 − r2 )
is relative phase difference due to path length difference.
λ
Note: φ is in radian
φ = 2π
P r1 r2 Tedious Method E p (t) = E cos(ω t + φ ) + E cos(ω t) = [E cos(ω t)cos φ − E sin(ω t)sin φ ] + E cos(ω t)
= [1+ cos φ ]E cos(ω t) − E sin φ sin(ω t) ≡ A cos(ω t) − Bsin(ω t)
A ≡ [1+ cos φ ]E,
B ≡ E sin φ
⎡
⎤
A
B
= A +B ⎢ 2
cos(ω t) −
sin(ω t) ⎥
2
2
2
A +B
⎣ A +B
⎦
A
B
B
sin φ
Define :
=
cos
δ
,
=
sin
δ
⇒
tan
δ
=
=
A [1+ cos φ ]
A2 + B2
A2 + B2
2
2
A 2 + B 2 = E [1+ cos φ ]2 + sin 2 φ = E 2(1+ cos φ )
⎛φ⎞
⎛φ⎞
= E 4 cos 2 ⎜ ⎟ = 2E cos ⎜ ⎟
⎝ 2⎠
⎝ 2⎠
B
sin φ
⎛φ⎞
⇒ E p (t) = 2E cos ⎜ ⎟ cos(ω t + δ ); tan δ = =
⎝ 2⎠
A [1+ cos φ ]
Phasor Diagram –Easy “recipe” Draw a vector with magnitude=E and
makes an angle =ω t wrt the horizontal axis
such that the horizontal component is Ecosω t.
Draw a vector with magnitude=E and
makes an angle =ω t+φ wrt the horizontal axis
such that the horizontal component is Ecos(ω t+φ ).
Add these two vectors. The resultant vector
will have the correct amplitude = 2E|cos
( )|
φ
2
sin φ
1+ cos φ
Equating the horizontal component of this vector to
the horzontal components of the two oringial vectors
provides the desired result:
and will have the correct angle =ω t+δ ; tan δ =
Ecosω t+Ecos(ω t+φ )=2E|cos
( ) | cos(ω t + δ )
φ
2
Use Euler’s IdenDty to explain the Phasor Diagram Method Euler's identity: eiθ = cosθ + i sin θ
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 cos 2θ + sin 2 θ = 1.
⇒ Eeiω t = E[cos ω t + i sin ω t] is the first vector on the phasor diagram
Eei(ω t+φ ) = E[cos(ω t + φ ) + i sin(ω t + φ )] is the second vector.
The resultant vector is the sum of these two vectors.
Q. What does the additon of these vectors, Eeiω t and Eei(ω t+φ ) ,
have to do with the Ecosω t + E cos(ω t + φ )?
See next slide.
The real part of (Eeiω t + Eei(ω t+φ ) ) = E cos ω t + E cos(ω t + 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:
Eeiω t + Eei(ω t+φ ) = Eeiω t [1+ eiφ ] = Eeiω t [1+ cos φ + i sin φ ]
Any complex number A+iB can be written as
its magnitude times an unit vector
A 2 + B 2 [cos δ + i sin δ ] = A 2 + B 2 eiδ ; tan δ =
B
A
⇒ [1+ cos φ + i sin φ ] = [1+ cos φ ]2 + sin 2 φ eiδ = [2 + 2 cos φ ]eiδ = 2 | cos φ2 | eiδ
tan δ =
sin φ
1+ cos φ
⇒ Eeiω t + Eei(ω t+φ ) = 2E | cos φ2 | eiω t eiδ
The real part is 2E | cos φ2 | cos(ω t + 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.