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WAVES
Optics
WAVE BEHAVIOR 3: DIFFRACTION
Diffraction is the bending of a wave AROUND a barrier
Diffracted waves can interfere and cause “diffraction
patterns”
DOUBLE SLIT DIFFRACTION

n = d sinΘ
n: bright band number
 : wavelength (m)
 d: space between slits (m)
 Θ: angle defined by
central band, slit, and
band n


This also works for diffraction gratings consisting of many,
many slits that allow the light to pass through. Each slit acts as
a separate light source
SINGLE SLIT DIFFRACTION

n  = s sin Θ
n: dark band number
 : wavelength (m)
 s: slit width (m)
 Θ: angle defined by central band, slit, and dark band

Light of wavelength 360 nm is passed through a diffraction
grating that has 10,000 slits per cm. If the screen is 2.0
m from the grating, how far from the central bright band
is the first order bright band?

Light of wavelength 560 nm is passed through two slits.
It is found that, on a screen 1.0 m from the slits, a bright
spot is formed at x = 0, and another is formed at x = 0.03
m. What is the spacing between the slits?
REFLECTION
Reflected sound can be heard as an echo
 Light waves can be drawn as “rays” to
diagram light reflected off mirrors

REFLECTION AND PLANE MIRRORS
 Law
of Reflection
MIRRORS
 Mirrors
can be
Plane (flat)
 Spherical

Convex – reflective side curves outward
 Concave – reflective side curves inward

OPTICAL IMAGES

Nature
Real (converging rays)
 Virtual (diverging rays)


Orientation
Upright
 Inverted


Size
True
 Enlarged
 Reduced

MIRRORS AND RAY TRACING
Ray tracing is a method of constructing an image using
the model of light as a ray
 We use ray tracing to construct optical images produced
by mirrors and lenses
 Ray tracing lets us describe what happens to the light as
it interacts with a medium

RAY TRACING: PLANE MIRRORS

Use at least two rays to construct the image
PROBLEM 4
 Standing
2.0 m in front of a small vertical mirror,
you see the reflection of your belt buckle, which
is 0.70 m below your eyes.


What is the vertical location of the mirror relative to
the level of your eyes?
If you move backward until you are 6.0 m from the
mirror, will you still see the buckle, or will you see a
point on your body that is above or below the
buckle? Explain.
SOLUTION
SPHERICAL MIRRORS
Concave
Convex
PARTS OF A
SPHERICAL CONCAVE MIRROR
+
-
Vertex
Center
Focus
Principle axis
PARTS OF A SPHERICAL CONCAVE MIRROR

The focal length is
half the radius of
curvature

R = 2f
The focal length is
positive for a
concave mirror
because it is on
the shiny side
 Rays parallel to
the optical axis all
pas through the
focus

RAY TRACING FOR CONCAVE MIRRORS
 You
must draw at least TWO of the three
principal rays to construct an image:
The p-ray: parallel to the principal axis, then reflects
through the focus
 The f-ray: travels through the focus then reflects back
parallel to the principal axis
 The c-ray: travels through the center, then reflects
back through the center

CONCAVE MIRRORS
Diagram the following images:

Object outside the center of curvature

Object at the center of curvature

Object between center of curvature and focus

Object at the focal point

Object inside the focus
SPHERICAL CONCAVE MIRROR
(OBJECT OUTSIDE CENTER)
c
p
C
f
F
Real,
Inverted,
Reduced
Image
SPHERICAL CONCAVE MIRROR
(OBJECT AT CENTER)
C
F
Real,
Inverted,
True
Image
SPHERICAL CONCAVE MIRROR
(OBJECT BETWEEN CENTER AND FOCUS)
C
F
Real,
Inverted,
EnlargedI
mage
SPHERICAL CONCAVE MIRROR
(OBJECT AT FOCUS)
C
F
No image
SPHERICAL CONCAVE MIRROR
(OBJECT INSIDE FOCUS)
C
F
Virtual,
Upright,
Enlarged
Image
MIRROR EQUATION 1 & MIRROR EQUATION 2
 1/si +
1/s0 = 1/f
si: image distance
 s0: object distance
 f: focal length

M
= hi/h0 = -si/s0
si: image distance
 s0: object distance
 hi: image height
 h0: object height
 M: magnification

SIGN CONVENTIONS

Focal length (f)
Positive for concave mirrors
 Negative for convex mirrors


Magnification (M)
Positive for upright images
 Negative for inverted images
 Enlarged: M > 1
 Reduced: M < 1



Practice: A spherical
concave mirror, focal
length 20 cm, has a 5-cm
high object placed 30 cm
from it


Image Distance
si is positive for real images
 si is negative for virtual images


Draw a ray diagram and
construct the image
Use mirror equations to
calculate the position,
magnification, and size of
the image
Name the image
Solution: A spherical concave mirror, focal length 20 cm,
has a 5-cm high object placed 30 cm from it.
a) Calculate the position, magnification, and size of the
image.
1 1 1
 
f si so
1
1
1
 
20 cm s i 30 cm
s i  60 cm
si
60
M  
 2
so
30
hi
M 
ho
hi  M ho  (  2)(5 cm )   10 cm
b) Name the image
real, inverted, enlarged.
PARTS OF A SPHERICAL CONVEX MIRROR
 The
focal length is
half the radius of
curvature and both
are on the dark side
of the mirror
 The focal length is
negative
RAY TRACING
Construct the image for
an object located outside
a spherical convex mirror
 Name the image

PRACTICE 5

A spherical concave mirror, focal length 15 cm, has a 4cm high object placed 10 cm from it
Draw a ray diagram and construct the image
 Use the mirror equations to calculate position, magnification,
and size of image
 Name the image

Problem: A spherical convex mirror, focal length 15 cm, has
a 4-cm high object placed 10 cm from it.
a) Use the mirror equations to calculate
i. the position of image 1
1 1
 
f si so
1
1
1
 
15cm si 10cm
ii.
the magnification
si  6cm
s

6
c
m
i
M


 

0
.
6
s
1
0
c
m
o
iii. the size of image
h

M
hc

(

0
.
6
)
(
4
m
)

2
.
4
c
m
i
o
b) Name the image
virtual, upright, reduced size
SPHERICAL MIRRORS
Concave
Convex
Image is real when object
is outside focus
 Image is virtual when
object is inside focus
 Focal length is positive

Image is ALWAYS virtual

Focal length is negative

REAL VS VIRTUAL IMAGES

Real
Formed by converging
light rays
 si is positive when
calculated with mirror
equation


Virtual
Formed by diverging
light rays
 si is negative when
calculated with mirror
equation

UPRIGHT VS INVERTED IMAGES

Upright
Always virtual if
formed by one mirror
or lens
 hi is positive when
calculated with
mirror/lens equation


Inverted
Always real if formed
by one mirror or lens
 hi is negative when
calculated with
mirror/lens equation

WAVE BEHAVIOR 2: REFRACTION
Refraction occurs when a wave is transmitted from one
medium to another
 Refracted waves may change speed and wavelength
 Refraction is almost always accompanied by some
reflection
 Refracted waves do NOT change frequency

REFRACTION OF LIGHT
Refraction causes a change in speed of light as it moves from
one medium to another
 Refraction can cause bending of the light ray at the interface
between media
 Index of Refraction (n):

n = speed of light in vacuum / speed of light in medium
 n = c/v

SNELL’S LAW
n1sinΘ1 = n2sinΘ2
SNELL’S LAW
When the index of refraction increases, light bends TOWARD
the normal
 When the index of refraction decreases, light bends AWAY
FROM the normal

PROBLEM 6!

Light enters an oil from the air at an angle of 50° with
the normal, and the refracted beam makes an angle of
33° with the normal
Draw the situation
 Calculate the index of refraction of the oil
 Calculate the speed of light in the oil

Solution: Light enters an oil from the air at an angle of 50o
with the normal, and the refracted beam makes an
angle of 33o with the normal.
a)
b)
c)
Draw this situation.
Calculate the index of refraction of the oil.
Calculate the speed of light in the oil
PRISM PROBLEMS (7)

Light enters a prism as shown, and passes through the prism
a. Complete the path of light through the prism and show the angle it
will make when it leaves the prism
b. If the refractive index of the glass is 1.55, calculate the angle of
refraction when it leaves the prism
c. How would the answer to (b) change if the prism were immersed
in water?
Solution: Light enters a prism as shown, and passes
through the prism.
a) Complete the path of the light through the prism, and show the
angle it will make when it leaves the prism.
b) If the refractive index of the glass is 1.55, calculate the angle of
refraction when it leaves the prism.
c) How would the answer to b) change if the prism were immersed
in water?
b)
n1 sin q1  n2 sin q 2
(1.55)sin(30)  (1.0)sin q 2
q 2  51
o
air
30o
q2
30o
c) In water, the angle of the light as it
leaves the glass would be smaller,
since the indices of refraction would be
more similar and there would be less
bending.
glass
60o
PRISM PROBLEMS (8)

Light enters a prism made of air from glass
a.
b.
Complete the path of the light through the prism, and show
the angle it will make when it leaves the prism
If the refractive index of the glass is 1.55, calculate the
angle of refraction when it leaves the prism
Problem: Light enters a prism made of air from glass.
a)
b)
Complete the path of the light through the prism, and show the
angle it will make when it leaves the prism.
If the refractive index of the glass is 1.55, calculate the angle of
refraction when it leaves the prism.
b)
n1 sin q1  n2 sin q 2
(1.00)sin(30)  (1.55)sin q 2
q 2  19o
30o
glass
q2
air
30o
60o
CRITICAL ANGLE OF INCIDENCE
The smallest angle of incidence for which light cannot
leave a medium is called the critical angle of incidence
 If light passes into a medium with a greater refractive
index than the original medium, it bends away from the
normal and the angle of refraction is greater than the
angle of incidence
 If the angle of refraction is ≥90°, the light cannot leave
the medium and no refraction occurs
 We call this TOTAL INTERNAL REFLECTION

TOTAL INTERNAL REFLECTION

Calculating the Critical
Angle:
PRACTICE NO. 9
What is the critical angle
of incidence for a
gemstone with refractive
index 2.45 if it is in air?
 If you immerse it in water
(refractive index 1.33),
what does this do to the
critical angle of
incidence?

Solution: What is the critical angle of incidence for a
gemstone with refractive index 2.45 if it is in air?
n1 sin q1  n2 sin q 2
(2.55)sin(q c )  (1.00)sin(90)
q c  23
o
If you immerse the gemstone in water (refractive index
1.33), what does this do to the critical angle of
incidence?
It increases the critical angle of incidence because there is
less difference in the refractive indices.
LENSES REFRACT LIGHT:
Converging (Convex)
Diverging (Concave)
LENS RAY TRACING

Ray tracing is used for lenses also. Use the same
principal rays used with mirrors. You must draw TWO of
the three:
the p-ray: parallel to the principal axis, refracts through the
focus
 the f-ray: travels through the focus, then refracts parallel to
the principal axis
 the c-ray: travels through the center and continues without
bending


Use the same equations we used for mirrors
CONVERGING LENSES
CONSTRUCT THE FOLLOWING IMAGES:
Object located outside 2F for a converging lens
 Object located at 2F
 Object located between F and 2F
 Object at the focus
 Object inside the focus

FOR CONVERGING LENSES
f is positive
• so is positive
• si is positive for real images and
negative for virtual images
• M is negative for real images and
positive for virtual images
• hi is negative for real images and
positive for virtual images
•
DIVERGING LENSES

Construct an image for an object located in front of a
diverging lens:
FOR DIVERGING LENSES
•f
is negative
• so is positive
• si is negative
• M is positive and < 1
• hi is positive and < ho