Download 2204-Unit4 – Part2 Notes

Snell’s Law
Amount of refraction (ie. angle change) depends on:
1.
2.
the index of refraction of the two media
the angle of incidence
This relationship is known as Snell’s Law:
n1sinθ1=n2sinθ2
i
n1sinθ1=n2sinθ2
r
ni
nr
where θ1 = the angle of incidence
θ2 = the angle of refraction
n1 = index of refraction for incident medium
n2 = index of refraction for refracting medium
Example 1:
A ray of light passes from air ( n=1.00) to water
(n =1.33), with an angle of incidence of 30°.
What is the angle of refraction?
na x sin 300 = nw x sin Ɵw
1.00 x ( 0.5) = 1.33 x sin Ɵw
0.5 = sin Ɵw
1.33
sin Ɵw = 0.3759
Ɵw = sin -1 ( 0.3759)
Ɵw = 220
Example 2:
Calculate the index of refraction for an incident
substance where the angle of incidence is 30°,
the angle of refraction is 50.0°, and the index of
refraction of the second substance is 1.50.
( n1 = 2.3 )
Review – don’t write out. Lets do together.
1. When light passes from a more optically dense medium
into a less optically dense medium, it will bend _______
(towards, away from) the normal.
2. When light passes from a medium with a high index of
refraction value into a medium with a low index of
refraction value, it will bend _______ (towards, away from)
the normal.
** The higher the value of n the more optically dense the
medium is, therefore the slower the light
moves.....causing the light ray to move towards the
normal when going from a high n value to a low one and
away from the normal when going from a low n value to a
higher one.
3. In each diagram, draw the "missing" ray (either
incident or refracted) in order to appropriately
show that the direction of bending is towards or
away from the normal.
4. Arthur’s method of fishing involves spearing
the fish while standing on the shore. The actual
location of the fish is shown in the diagram
below. Because of the refraction of light, the
observed location of the fish is different than its
actual location. Where does the fish appear to
be? Must Arthur aim above or below where the
fish appears to be in order to strike the fish?
5. A ray of light in air is approaching the boundary
with a layer of crown glass at an angle of 42.0
degrees. Determine the angle of refraction of the
light ray upon entering the crown glass and upon
leaving the crown glass.
1 sin 42 = n2 sin Θ2 = 1 sin Θ3
Θ3 = 42
Practice: P.438 #’s 64, 66, 69, 60
a)
b)
c)
Θ1 = 45o
Θ1 = 48o
Θ1 = 49o
Comment on what is happening to the refracted light ray.
Critical Angle
nr
ni
c
The critical angle for a particular material is the
angle of incidence for which the angle of
refraction is 90o.
The symbol for critical angle is Θc.
In order for a critical angle to exist a material
must always be optically denser than the 2nd
material. ie. n1 > n2

From Snell,
n1 sinc = n2 sin 90
Since sin 90 = 1, we have
n1 sinc = n2
and the critical angle is
c =
sin-1 nn
2
i
Example:
What is the critical angle of a block of crystal
glass (n = 1.54) sitting in air (n = 1).
Don't forget, material 1 is the glass.
Answer: 40.5o
What is the point of knowing this?..........
Some very useful inventions have resulted by
taking advantage of total internal reflection....
1. Periscope (tanks, submarines,...)
2. Binoculars
◦ uses a class prism to reflect light to prevent loss of
light (pg.417)
3. Endoscopes (medical camera)
◦ allows doctor to explore a patient internally without
unnecessary surgery. (pg. 417)
4. Bike reflectors (pg.417)
Assign: P.439 # 79, 80.
If you were to shine a laser (light of one color) at a barrier with a
tiny slit (opening), what would you see on the screen?
A) a single line of light
B) a central light that gradually gets dimmer as you move out
C) an alternating series of bright and dark lines that
gradually gets dimmer as you move out
Answer: C)
The pattern’s size and location varies with the width of the slit.
When light passes through a single opening
that is close in size to it’s wavelength, the
light waves interact forming a diffraction
pattern.






Light passes through
a single opening of
width w.
A central bright spot
is created, at center.
An alternating dark
then bright pattern
develops on both
sides.
We use a formula to
locate dark spots
n=1, 2, 3, 4, 5 ...
We could locate
bright spots at
n=1.5, 2.5, 3.5...
Formula for minima, ie. dark spots
n  wsin 
n  wsin 
n=nodal line # (minima/darkspot: n=1,2,3...)
λ = wavelength of light (m)
w= width of slit (m)
θ = angle away from center line
(degrees)
Example
A 20.0 µm slit is illuminated by a red light (620 nm). At what
angle is the third order minimum located ?
w = 20.0 x 10
nλ=wsinθ
-6
m
λ = 620 x 10
θ = Sin -1 (nλ/w)
θ = Sin -1 (3 x 620 x 10
θ = 5.30o
-9/
-9
20.0 x 10
m
-6
n=3
)
Assign p. 441 # 100 & 101.
Light’s Diffraction Pattern:
Wave, particle or
something else ?
Watch

https://www.youtube.com/watch?v=Iuv6hY6
zsd0#t=403
Young’s Double Slit Experiment
When light passes through 2 small slits with a small separation (d) between
slits they may create an interference pattern. For light, crests and troughs
produce bright and dark spots on a screen.
nλ = dsinθn
n= order number (bright spot from center line)
λ = wavelength of light
d= slit separation (distance between 2 slits)
θn = angle from center to nth maxima
n= order number
dxn
n 
L
λ = wavelength of light
d= slit separation (distance between 2 slits)
xn =distance from center to nth maxima
L = distance from centre of slits to maxima
Ex. 1:
A 450 nm light passes through two slits with separation of 3.00 µm.
a) Find the angle at which the second order maximum is located on
a screen 1.00 m from the slits.
n=2
λ = 450 nm = 450 x 10 -9 m
d = 3.00 x 10 -6 m
θ2 = ?
nλ = dsinθn
θ2 = sin-1 (n λ/d)
= 17.5o
b) How far from the centre line is this second maximum located?
X = 0.300 m or 30.0 cm
** How would you find Minima (dark spots)? n = 0.5, 1.5, 2.5 ...
Practice p. 440 #92-94.

Speed
– depends on medium.
 Sound in Air – temperature, pressure, etc.
 Speed of Sound in Air vsound = 332m/s + 0.6 T
Note: T is temperature in oC
 Waves in string – tension, string diameter/mass

Frequency – depends on source
 Wave’s Frequency = Frequency of vibrating source

Wavelength (λ) – depends on Eq’n: λ = v/f

https://www.youtube.com/watch?v=imoxDcn
2Sgo

The frequency of a wave does not change
once created. However, if the source and
observer are moving relative to each other
there is an apparent change in frequency, a
Doppler shift.

http://www.absorblearning.com/advancedphysics/
demo/units/040103.html#Introduction
Note:


Doppler effect involves apparent change in
frequency observed due to relative velocity and not
the change in loudness due to distance.
Doppler effect for sound:
f1vs
f2 =v ± v
s
o
Where:
f2 = apparent frequency (Hz)
f1 = original frequency of source (Hz)
vs = speed of sound (m/s)
vo = relative speed of observer (m/s)
+ when source/observer move apart
- when source/observer move towards each other
Ex.
It is 10.0 oC and you are moving toward a
stationary alarm which has a frequency of
925 Hz. What is the apparent frequency you
hear if you are traveling at 101 km/h ?
Answer ( 1010Hz )
Assign P. 480 # 64-66.
Water vapour
condensed due
to Sonic Boom.

If the source travels faster than the wave it
will pass it’s own wave causing new waves to
overlap the previous ones. This gives
constructive interference at a V shape behind
Shock wave behind boat moving > speed of water wave
the object.
An apparent shift in frequency can also occur with
light waves. The difference in these frequencies tells
us the speed of the object.
Main Applications:
 Radar guns
 Astronomy
 Weather Radar
Use the following for radar guns:
Where f   f  f 
2
1
f2 is the measured frequency,
f1 is the emitted frequency
r
vr = relative velocity of object
1
c = speed of light
 f
v  
2f

c

Ex. A stationary police officer uses a radar gun emitting EM
waves at 9.00 x 10 9 Hz. The detected waves differ by
2000 Hz. What is the speed of the approaching car?
 f 
c
vr  
 2 f1 
2000 

8
vr  
3
.
00
x
10

9
 2 x9.00 x10 
= 33 m/s
In astronomy the difference in colors (λ or f) is compared
when a star or galaxy is moving. The change or shift in
color tells us how fast it’s traveling.
Where:
f2 is the observed frequency (Hz)
f1 is the source frequency (Hz)
2
1
v = relative velocity of object
c = speed of light (3.00x108m/s)
 v
f  f 1  
 c
Ex. Star ‘Swerc’ normally emits a frequency of 5.77x1014 Hz.
On Earth we observe 5.76x1014 Hz. What is the speed of
the star and is it moving toward or away from us?
Lower f = moving away (use – sign)
Answer: (520km/s)
In Example 1 the star is
going away from the source.
This is called a red shift
because the wavelength has shifted toward
the red end of spectrum.
The opposite would be a blue shift.
Practice P. 440 # 83,84,87