Download 4.1 The Concepts of Force and Mass

Lenses and the Refraction of
Light
26.1 The Index of Refraction
• Light travels through a vacuum at at c
c  3.00 108 m s
• Light travels through materials such as water or glass at a speed
less than c.
• Light will bend, or refract, if it enters a medium with a different
speed. It may reflect as well.
• The amount of bending is quantified by 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. As a ratio, it is unitless:
n
Speed of light in vacuum
c

Speed of light in the material v
26.2 Snell’s Law and the Refraction of Light
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 1  n2 sin 2
where n1 is the index of refraction in the
incident ray medium, and n2 is that
index of refraction in the refracting ray
medium.
There is also a reflected ray at an angle
of θ1.
A mirror surface does not refract, it only
reflects.
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.
From air to water
n1 = 1
θ1 = 46°
n2 = 1.33
Find θ2
From water to air
n1 = 1.33
θ1 = 46°
n2 = 1
Find θ2
n1 sin 1  n2 sin 2
26.2 Snell’s Law and the Refraction of Light



n
sin

1
.
00
sin
46
1
1
(a) sin  2 

 0.54
n2
1.33
 2  sin 1 (.54)  33
When a ray is transmitted into a material
with a higher index of refraction it bends to
make a smaller angle with the normal.
(b)
n1 sin 1 1.33sin 46
sin  2 

 0.96
n2
1.00
 2  sin 1 (.96)  74
When a ray is transmitted into a material
with a lower index of refraction it bends to
make a larger angle with the normal.
26.2 Snell’s Law and the Refraction of Light
APPARENT DEPTH – when an object is in a medium with a different
index of refraction that that of the observer, the observer will see a
virtual image of the object which is at a different location than the
object
Apparent depth,
observer directly
above object
Where n1 is for the
medium of the
incident ray
coming from the
object and n2 is
for the medium of
the refracted ray
 n2 
d   d  
 n1 
Workbook exercises
26.2 Snell’s Law and the Refraction of Light
Conceptual Example 4 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. To the
swimmer, the coin appears to be at a certain height above the
water. Is the apparent height of the coin:
A. Greater than its actual height
B. Less than its actual height
C. Equal to its actual height
26.2 Snell’s Law and the Refraction of Light
THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL
Incident ray and emergent ray (doubly refracted) are equal.
A light ray
travels from
medium 1 to
medium 3 as
shown.
For these
media,
A. n3 = n1.
B. n3 > n1.
C. n3 < n1.
D.We can’t compare n1 to n3 without knowing n2.
26.3 Total Internal Reflection
When light passes from a medium of larger refractive index into one
of smaller refractive index, the refracted ray bends away from the
normal.
At some point, the incident angle will be such that the reflected angle is
90° from the normal, i.e. no part of the ray will refract out of the 2nd
medium.
That incident angle is the incident angle, θc.
Recall sin 90° = 1
Critical angle
n2
sin  c 
n1
n1  n2
26.3 Total Internal Reflection Fiber Optics
The glass core of an optical fiber has an
index of refraction of 1.60. The cladding has
an index of refraction of 1.48. What is the
maximum angle a light ray coming from the
air can make with the core wall if it is to
remain inside the fiber?
26.3 Total Internal Reflection Fiber Optics
The glass core of an optical fiber has an
index of refraction of 1.60. The cladding has
an index of refraction of 1.48. What is the
maximum angle a light ray coming from the
air can make with the core wall if it is to
remain inside the fiber?
Known
ncore = 1.60
nclad = 1.48
Find
θ1
n1 sinθ1 = n2 sin θ2 Law of refraction
θ2 = 90 – θc Geometry – see illustration
sin θc = nclad /ncore Total internal reflection
26.3 Total Internal Reflection Fiber Optics
The glass core of an optical fiber has an
index of refraction of 1.60. The cladding has
an index of refraction of 1.48. What is the
maximum angle a light ray coming from the
air can make with the core wall if it is to
remain inside the fiber?
Known
ncore = 1.60
nclad = 1.48
Find
θ1
nair θ1 = ncore θ2 Law of refraction
θ2 = 90 – θc Geometry – see illustration
sin θc = nclad /ncore Total internal reflection
θ1 =37.44 degrees.
26.5 The Dispersion of Light: Prisms and Rainbows
The net effect of a prism is to change the direction of a light ray.
Light rays corresponding to different colors bends by different amounts.
26.5 The Dispersion of Light: Prisms and Rainbows
A narrow beam of sunlight is incident on a slab of
crown glass at 45 degrees. Find the angle
between the violet ray and the red ray in the
glass.
26.5 The Dispersion of Light: Prisms and Rainbows
A narrow beam of sunlight is incident on a slab of
crown glass at 45 degrees. Find the angle
between the violet ray and the red ray in the
glass.
Answer: 0.35 degrees
26.5 The Dispersion of Light: Prisms and Rainbows
26.6 Lenses
•Mirrors reflect light according to the Law of Reflection and
create a real or virtual image of the light source.
•Lenses refract light according to Snell’s Law of Refraction
and create a real or virtual image of the light source.
•Converging and diverging lenses are analogous to
concave and convex mirrors.
26.6 Lenses
With a converging lens, paraxial rays that are parallel
to the principal axis converge to the focal point of the
lens on the opposite side from the object.
26.7 The Formation of Images by Lenses
Ray diagrams for converging lenses:
•Ray 1 is parallel when it leaves the object and is
refracted so that it passes through the focal point on the
other side of the lens.
•Ray 2 passes through the focal point on the same side
as the object and is refracted so that it is parallel on the
other side of the lens.
•Ray 3 passes though the principal axis in the middle of
the lens and continues on in a straight line.
26.7 The Formation of Images by Lenses
There are 3 different types of images formed by a
converging lens, depending on the relationship between the
object and the focal point of the lens.
Object more than twice the focal length from
the lens
• In this example, when the object is placed further than
twice the focal length from the lens, the real image is
inverted and smaller than the object.
• When using a digital camera, the film is replaced by a
charge coupling device (CCD), a semi-conductor diode
that acts much like a solar cell and changes light into
electrical current.
Object further away than focal length
• When the object is placed between F and 2F, the real image is
inverted and larger than the object.
• To see the real image from a projector on the screen, you have
to be between the lens and the screen and look at the reflection
from the screen
26.7 The Formation of Images by Lenses
IMAGE FORMATION BY A CONVERGING LENS (MAGNIFYING GLASS)
2
• When the object is placed between F and the lens, the virtual image
is upright and larger than the object.
• Now you look through the lens, but you don’t see the object itself, but
rather it’s virtual image.
•Note the lens is not high enough for #2 , but I drew it in.
26.6 Lenses
With a diverging lens, paraxial rays that are parallel to
the principal axis appear to originate from the focal
point on the same side as the object.
26.7 The Formation of Images by Lenses
Ray diagrams for diverging lenses:
•Ray 1 is parallel when it leaves the object and is
refracted so that it’s continuation passes through the
focal point on the same side of the lens as the object.
•Ray 2 is refracted so that it is parallel and its
continuation passes through the focal point on the
opposite side of the lens.
•Ray 3 passes though the principal axis in the middle of
the lens and continues on in a straight line.
26.7 The Formation of Images by Lenses
IMAGE FORMATION BY A DIVERGING LENS
2
•A
diverging lens always forms an upright, virtual,
reduced image.
• Ray 2 leaves the object and travels toward the
focal point on the other side of the lens.
A lens produces a sharplyfocused, inverted image on a
screen. What will you see on
the screen if the lens is
removed?
A.
B.
C.
D.
E.
The image will be as it was, but much dimmer.
The image will be right-side-up and sharp.
The image will be right-side-up and blurry.
The image will be inverted and blurry.
There will be no image at all.
26.8 The Thin-Lens Equation and the Magnification Equation
1 1 1
 
do di
f
hi
di
m 
ho
do
26.8 The Thin-Lens Equation and the Magnification Equation
Summary of Sign Conventions for Lenses
f is  for a converging lens.
f is  for a diverging lens.
d o is  for all cases studied in this class.
di is + for a real image formed on the opposite side of the lens as the
object.
di is - for a virtual image formed on the same side of the lens as the
object.
m is  for an upright image.
m is  for an inverted image.
A converging lens
A 20 cm high object is located 30.0 cm to the
left of a converging lens whose focal length is
50.0 cm. What is the nature of the image?
A.
B.
C.
D.
real, inverted and reduced
real, inverted and enlarged
virtual, upright and enlarged
virtual, upright and reduced.
A converging lens
A 20 cm high object is located 30.0 cm to the
left of a converging lens whose focal length is
50.0 cm. What is the a) position and
b)magnification of the image?
Solve for di and m
Converging lens
From the thin-lens equation, we obtain:
(1/ do )  (1/ di )  (1/ f )
1 1 1
1
1
 –

–
d i f d o 50.0 cm 30.0 cm
or
d i = –75.0 cm
The negative answer shows it is on the same side of the lens as
the object. The magnification equation gives the magnification
to be:
di
–75.0 cm
m–
–
 +2.50
do
30.0 cm
The positive number shows it is upright.
A converging lens
A 20 cm high object is located 30.0 cm to the
left of a converging lens whose focal length is
50.0 cm. What is the a) position and
b)magnification of the image?
Solve for di and m
26.10 The Human Eye
Anatomy of the Eye
•The cornea and the lens act
to bend the incoming light
waves.
• The lens only contributes
about 20-25% of the
refraction, but its function
is important, as it provides
the fine-tuning mechanism.
• The retina acts as the film,
or CCD and receives the
image, which it then sends,
via the optic nerve, to the
brain, which interprets the
image.
26.10 The Human Eye
OPTICS
•When the ciliary muscles relax, the lens
becomes less curved.
•The radius of curvature, and therefore the focal
length increases.
•The eye can focus on more distant objects.
•Of course there is a limit to how far away an
object can be that can be focused at the retina,
even for people with good vision.
26.10 The Human Eye
OPTICS
•When the ciliary muscles become tensed, the lens
becomes more convex.
•The radius of curvature, and therefore the focal
length, decreases.
• This allows the eye to focus on closer objects.
• Of course there is a limit to this close focus.
26.10 The Human Eye
NEARSIGHTEDNESS
• For some people, the lens cannot become “relaxed” enough
to focus far-away objects on the retina. Instead, the eye
focuses the image in front of the retina.
•For the purposes of this discussion, “a distant” object is
considered to be infinitely far away (d0 = ∞).
26.10 The Human Eye
NEARSIGHTEDNESS
A divergent lens in front of the eye creates a virtual image in
front of the object, close enough for the nearsighted eye to
see.
The virtual image acts as a virtual object, from which the eye
makes a real image on the retina.
26.10 The Human Eye
FARSIGHTEDNESS
•The lenses of some people cannot tense up enough
to focus on a close object.
•The lens creates an image of the close object behind
the retina.
• This happens to all people as they age !!
•Now the objective is to create a virtual image further
away than the object.
26.10 The Human Eye
FARSIGHTEDNESS
The lens creates a virtual image of the close object at the near point
of the farsighted eye.
26.10 The Human Eye
Example 12 Eyeglasses for the Nearsighted Person
A nearsighted person has a far point located only 521 cm from the
eye. Assuming that eyeglasses are to be worn 2 cm in front of the
eye, find the focal length needed for the diverging lens of the glasses
so the person can see “distant” objects.
26.10 The Human Eye
d0 = ∞
di = (521-2)= 519 cm
1
0

Find
f
1
1 1


f do di
1
1 1 1
1

  
f d o d i  519 cm
f  519 cm
26.10 The Human Eye
THE REFRACTIVE POWER OF A LENS – THE DIOPTER
Optometrists who prescribe correctional lenses and the opticians
who make the lenses do not specify the focal length. Instead
they use the concept of refractive power.
1
Refractive power (in diopters) 
f in meters 
26.11 Angular Magnification and the Magnifying Glass
The size of the image on the retina determines how large
an object appears to be.
26.11 Angular Magnification and the Magnifying Glass
ho
 in radians   Angular size 
do
26.11 Angular Magnification and the Magnifying Glass
Example 14 A Penny and the Moon
Compare the angular size of a penny held at arms length with that of
the moon.
Penny
ho 1.9 cm
 
 0.027 rad
d o 71 cm
Moon
ho 3.5 106 m
 
 0.0090 rad
8
d o 3.9 10 m
26.11 Angular Magnification and the Magnifying Glass
Angular magnification

M

Angular magnification
of a magnifying glass
1 1
M     N
 f di 
26.12 The Compound Microscope
To increase the angular magnification
beyond that possible with a magnifying
glass, an additional converging lens
can be included to “premagnify” the
object.
Angular magnification of
a compound microscope

L  f e N
M 
fo fe
26.13 The Telescope
Angular magnification of
an astronomical telescope
M 
fo
fe