Physics learning objective

1. Shawn Zhang
LC2

2.  When waves propagate from 2
sources, and come in contact with
each other, interference occurs
 It is a 2D representation of the
interference that arise from waves
overlapping

3.  2 sound source placed next to each
other
 Other signals distorting your WiFi

4.  Constructive
 When waves overlap their peaks with
peaks, and troughs with troughs
 Destructive
 When waves overlap peaks with
troughs, and vice versa

6.  As you can see from the previous
picture, the waves overlapped
 The graph above the waves show
areas of constructive and destructive
interference
 To overlap peaks with peaks and
troughs with troughs, the phase
difference must be 2
 Destructive interference is caused by
phase difference of

7.  Also, we can look at only one
portion of the wave to determine the
interference at a specific point in
space


8.  To determine the path difference
and interference pattern at point P
on the previous slide, we can solve it
by using trigonometry
 If the distance and/or angle is given,
we will solve for the unknown and
after obtaining the distances, we will
divide by the wavelength to see the
phase difference caused by path
difference is

9.  In this example, the phase difference can be
seen from the graph at the overlapping point
 If only numerical info were given, we will need
to solve it via trig
Practice time!

10.  Using the same pic as a reference, the angle
that wave 1 is 60 degrees from the slit
 Wave 2’s slit is 10m below wave 1
 Point P is 50m away from the wall where the
slits are
 The wave length of the waves are 5m
 What kind of interference occurs at point P?

11.  The information that we have are that S1 and
S2 are 10m apart, while P is 50m away from the
wall, and also that wave 1 propels at a 30
degree angle
 From the info we are given
we can make a new pic

12.  Using trig, we can determine that the distance
travelled by Wave 1 is 100m, and the vertical
distance from S1 to point P is 86.60m
 The vertical distance from S2 to point P will be
86.60 – 10 = 76.60m
 Once the vertical distance from S2 to point P is
obtained, we can calculate the distance
travelled by wave 2
 sqrt(76.60^2 + 50^2) = 91.4744m
 91.4744/5 = 18.2949 wavelengths
 100/5 = 20.00 wavelengths
 20-18 = 2 wavelengths apart = Constructive

13.  From the calculations on the previous page, it
can be seen that at point P, there is constructive
interference, because the path difference is an
integer number of wavelengths a part
 That concludes my LO for this week