Final Exam information
•  Wednesday, June 6, 2012, 9:30 am - 11:18 am
•  Location: in recitation room
•  Comprehensive (covers all course material)
•  35 multiple-choice questions --> 175 points
•  Closed book and notes
•  Make up your own equation sheet (same rules as midterm)
Chapter 25
The Reflection of
Light: Mirrors
25.5 The Formation of Images by Spherical Mirrors
IMAGING WITH CONCAVE MIRRORS
To find the image of an object placed in front of a concave mirror, there are
several types of rays which are particularly useful è ray tracing:
This ray is initially parallel to the principal axis
and passes through the focal point.
This ray initially passes through the focal point,
then emerges parallel to the principal axis.
This ray travels along a line that passes through
the center and so reflects back on itself.
25.5 The Formation of Images by Spherical Mirrors
If the object is placed between F and C, the image is real, inverted and
magnified. If the object is placed at a distance greater than C from the mirror,
the image is real, inverted and reduced in size. A real image is one where
light is actually passing through the image (it can be projected onto a screen).
The principle of reversibility - If the direction of a light ray is reversed,
the light retraces its original path.
25.5 The Formation of Images by Spherical Mirrors
When an object is placed between the focal point F and a concave mirror,
The image is virtual, upright, and magnified (as in the case of images
from flat mirrors, a virtual image is one from which light appears to be
emanating but through which light does not pass, e.g. it cannot be projected
onto a screen).
25.5 The Formation of Images by Spherical Mirrors
IMAGING WITH CONVEX MIRRORS
For convex mirrors the image of an object is
always virtual, upright, and reduced in size.
Ray 1 is initially parallel to the principal axis and appears to originate from
the focal point.
Ray 2 heads towards the focal point, emerging parallel to the principal axis.
Ray 3 travels toward the center of curvature and reflects back on itself.
25.6 The Mirror Equation and Magnification
So far we have discussed concave and convex mirrors qualitatively
and graphically. We now want to derive two simple equations which
provide quantitative relationships among the quantities we have
defined to describe mirrors, i.e.,
f = focal length
d o = object distance
d i = image distance
m = magnification
25.6 The Mirror Equation and Magnification
Consider the real image produced from a concave mirror.
The two right triangles are similar in each case.
ho/(-hi) = do/di minus since inverted
hi
di
m= =−
ho
do
Magnification
equation
ho/(-hi) = (do - f )/f
1 1 1
+ =
do di f
Mirror equation
Both equations are valid for concave and convex mirrors.
25.6 The Mirror Equation and Magnification
Summary of Sign Conventions for Spherical Mirrors
f is + for a concave mirror.
f is − for a convex mirror.
d o is + if the object is in front of the mirror.
d o is − if the object is behind the mirror.
d i is + if the image
object is in front of the mirror (real image).
d i is − if the image
object is behind the mirror (virtual image).
m is + for an image upright with respect to the object.
m is - for an image inverted with respect to the object.
Example. A Real Image formed by a Concave Mirror.
A 2.0 cm high object is placed 7.10 cm from a concave mirror whose
radius of curvature is 10.20 cm. Find the location of the image and its size.
Since a concave mirror è f = +R/2 = 10.20/2 = 5.10 cm
1/di = 1/f - 1/do = 1/5.10 - 1/7.10 = 0.055 cm-1
di = 18 cm è real image since positive
hi = -(di/do)ho = -(18/7.10)(2.0) = -5.1 cm è magnified and inverted
25.6 The Mirror Equation and Magnification
Example. A Virtual Image Formed by a Convex Mirror
A convex mirror is used to reflect light from an object placed 66 cm in
front of the mirror. The focal length of the mirror is 46 cm in back of the
mirror. Find the location of the image and the magnification.
Since a convex mirror, the focal length is negative è
f = -46 cm
1 1 1
1
1
= − =
−
= −0.037 cm −1
d i f doi − 46 cm 66 cm
d i = −27 cm
è image is virtual since negative
(
di
− 27 cm )
m=− =−
= 0.41
do
66 cm
è image is upright
and reduced
Chapter 26
The Refraction of
Light: Lenses and
Optical Instruments
26.1 The Index of Refraction
8
Light travels through a vacuum at a speed c = 3.00 × 10 m s
Light travels through materials at a speed less than its speed
in a vacuum.
DEFINITION OF THE INDEX OF REFRACTION
The index of refraction of a material is the ratio of the speed
of light in a vacuum to the speed of light in the material:
n=
Speed of light in vacuum
c
=
Speed of light in the material v
26.1 The Index of Refraction
26.2 Snell’s Law and the Refraction of Light
SNELL’S LAW -- When light strikes an interface between two materials it
breaks up into two pieces - one reflected and one refracted (transmitted).
SNELL’S LAW OF REFRACTION
When light travels from a material with
one index of refraction to a material with
a different index of refraction, the angle
of incidence is related to the angle of
refraction by
n1 sin θ1 = n2 sin θ 2
26.2 Snell’s Law and the Refraction of Light
Example 1 Determining the Angle of Refraction
A light ray strikes an air/water surface at an
angle of 46 degrees with respect to the
normal. Find the angle of refraction when
the direction of the ray is (a) from air to
water and (b) from water to air.
26.2 Snell’s Law and the Refraction of Light
n1 sin θ1 (1.00 )sin 46
(a) sin θ 2 =
=
= 0.54
n2
1.33
θ 2 = 33
(b)
n1 sin θ1 (1.33)sin 46
sin θ 2 =
=
= 0.96
n2
1.00
θ 2 = 74
26.2 Snell’s Law and the Refraction of Light
APPARENT DEPTH
Example 2 Finding a Sunken Chest
The searchlight on a yacht is being used to illuminate a sunken
chest. At what angle of incidence should the light be aimed?
26.2 Snell’s Law and the Refraction of Light
First find θ2 from the geometry
and then use Snell’s Law to find θ1:
θ 2 = tan −1 (2.0 3.3) = 31
n2 sin θ 2 (1.33)sin 31
sin θ1 =
=
= 0.69
n1
1.00
θ1 = 44
26.2 Snell’s Law and the Refraction of Light
Because light from the chest is
refracted away from the normal
when the light enters the air, the
apparent depth of the image is less
than the actual depth.
Simpler case -- look directly above
the object.
Apparent depth,
observer directly
above object
& n2 #
d ' = d $$ !!
% n1 "
n1 -- medium of object n2 -- medium of observer
26.2 Snell’s Law and the Refraction of Light
Example. On the Inside Looking Out
A swimmer is under water and looking up at the surface. Someone
holds a coin in the air, directly above the swimmer’s eyes at a distance
of 50 cm above the water. Find the apparent height of the coin as seen
by the swimmer (assume n = 1.33 for water).
Use the equation
& n2 #
d ' = d $$ !!
% n1 "
In this case, d’ will be the apparent height of the coin, d is the actual
height above the water, n1 = 1.00 for air (object), and n2 = 1.33 for
water (the observer),
d’ = (50)(1.33/1.00) = 66.5 cm è greater than the actual height
26.2 Snell’s Law and the Refraction of Light
THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL
When a ray of light passes
through a pane of glass that
has parallel surfaces and is
surrounded by air, the
emergent ray is parallel to
the incident ray, θ3 = θ1,
but is displaced from it.
1st interface: n1 sin θ1 = n2 sin θ2
2nd interface: n2 sin θ2 = n1 sin θ3
è n1 sin θ1 = n1 sin θ3 è θ3 = θ1