Information for Physics 1201 Midterm 2 Wednesday, March 27

My lecture slides are posted at
http://www.physics.ohio-state.edu/~humanic/
Information for Physics 1201 Midterm 2
Wednesday, March 27
1) Format: 10 multiple choice questions (each worth 5 points) and
two show-work problems (each worth 25 points), giving 100 points
total.
2) Closed book and closed notes.
3) Equations and constants will be provided on the midterm
4) Covers the material in Chapters 21, 22, 23, and 24

Polarization
Polarized light is produced by the scattering of unpolarized
sunlight by molecules in the atmosphere.
Molecules re-radiate sunlight.
As you look at larger and larger
angles with respect to the
incident sunlight, the re-radiated
light becomes more and more
horizontally polarized.

Polaroid Sun Glasses
Besides sunlight re-radiated from atmospheric molecules, there are
other sources of horizontally polarized light that occur in nature, such
as sunlight reflected from horizontal surfaces such as lakes.
Polaroid sun glasses take advantage of this fact by using polarizers
with their axes oriented vertically:
è in addition to 1/2 of the unpolarized light which is already blocked by
the polarizers, all of the horizontally polarized light is completely blocked,
thus blocking out some of the reflected light which can confuse you
during some outdoor activities (e.g. driving, piloting, fishing…….).

Polarization
Another application using polarized glasses: watching 3-D movies.
In a 3-D movie, two separate rolls of film are projected using a projector
with two lenses, each with its own polarizer. The two polarizers are
crossed. Viewers watch the action on-screen through glasses that have
correspondingly crossed polarizers for each eye.
IMAX movie projector Movie viewer using polarized glasses

Example. Partially polarized and partially unpolarized light.
A light beam passes through a polarizer whose transmission axis makes
angle θ with the vertical. The beam is partially polarized and
partially unpolarized, and the average intensity of the incident light,
S 0 , is the sum of the average intensity of the polarized light, S 0,polar ,
and the average intensity of the unpolarized light, S 0,unpolar .
As the polarizer is rotated clockwise, the intensity of the transmitted light
has a minimum value of 2.0 W/m 2 when θ = 20.0 o and has a maximum
value of 8.0 W/m 2 when the angle is θ = θ max .
è Find a) S 0,unpolar , and b) S 0,polar .
Incident light
S 0 = S 0,polar + S 0,unpolar
Transmitted light
S = S polar + S unpolar

Incident light
S 0 = S 0,polar + S 0,unpolar
Transmitted light
S = S polar + S unpolar
a) Minimum transmitted intensity S = S min = 2.0 W/m 2 at θ = 20.0 o .
S is minimum when S polar = 0 since S unpolar is not effected by θ.
S unpolar = S min - S polar = 2.0 - 0 = 2.0 W/m 2
S unpolar = ½ S 0,unpolar è S 0,unpolar = 2S unpolar = 2(2.0) = 4.0 W/m 2
b) Maximum transmitted intensity S = S max = 8.0 W/m 2 occurs at θ max .
S is maximum when S polar = S 0,polar (when θ = 20 o + 90 o = 110 o = θ max )
S max = S 0,polar + S unpolor è S 0,polar = S max - S unpolar = 8.0 - 2.0 = 6.0 W/m 2

Polarization and the Reflection and Refraction of Light
For air/water
interface:
θ p
θ p
n 1 =1.00, n 2 =1.33 ⇒ θ p = tan −1 1.33
n 1 sinθ p = n 2 sinθ 2 ⇒ θ 2 = sin
θ p +θ 2 = 90 o à Always true at θ p
Reflected light is
100% horizontally
polarized at the
polarizing angle, θ p
Brewster’s law
tanθ p = n 2
n 1
# &
% ( = 53.1
$ 1.00 '
o
−1 ( 1.00)sin53.1
o
)
,
+
. = 36.9
* 1.33 -
o

Chapter 25
Optical Instruments
and the Eye

Cameras
IMAGE FORMATION BY A CONVERGING LENS IN A CAMERA
When the object is placed further than twice the focal length
from the camera lens, the real image is inverted and smaller
than the object on the film or ccd chip.

Cameras
Example: Taking a picture of a flower vase with a ccd camera.
A 0.500 m high flower vase is positioned 2.00 m in front of a ccd camera.
The camera uses a converging lens whose focal length is 60.0 mm.
(a) How far must the ccd be from the lens to have a sharp image?
(b) Find the height of the image on the ccd.
(c) If the lens diameter is 3.0 cm, find the f-stop of the lens.
(d) If a properly exposed picture is taken with this f-stop with a shutter
speed of 1/500 s, what f-stop is needed for a shutter speed of 1/125 s
so that the picture is not overexposed?
(a)
(b)
1
d i
= 1
f
− 1
d o
=
1
0.0600 m −
d i = 0.0619 m = 61.9 mm
h i = − d i
d o
h o = − 0.0619
2.00
1
2.00 m
=16.17 m−1
0.500 = −0.0155 m = −15.5 mm

(c)
f − stop =
= 60 mm
f
Diameter of lens opening
30 mm
= 2.0 ⇒ f / 2
= f
D
(d) For the picture not to be overexposed, we must reduce the lens area
by the ratio of the shutter speeds, i.e.
A final =
!
#
"
π D final
2
1 500
1 125 Ainitial = 1
4 Ainitial $
&
%
2
= 1
f − stop final =
!
#
4 π Dinitial " 2
f
D final
= 2
$
&
%
2
f
D initial
⇒ D final = 1
2 D initial
= 2( 2.0)
= 4.0 ⇒ f / 4

The Human Eye
ANATOMY

The Human Eye
OPTICS
The lens only contributes about 20-25% of the refraction, but its function
is important.
Normal eye near point, i.e. nearest point for comfortable focusing, N = 25 cm
Normal eye far point, i.e. farthest point for comfortable focusing à ∞

The Human Eye
NEARSIGHTEDNESS
The lens creates an image of the distance object at the far point
of the nearsighted eye.

The Human Eye
Example: Eyeglasses for the Nearsighted Person
A nearsighted person has a far point and near point located at only 10.0 cm
and 8.0 cm, respectively, from the eye. Assuming that eyeglasses are to be
worn 2.0 cm in front of the eye, find (a) the focal length needed for the lens
of the glasses so the person can see distant objects and (b) the resulting
location of the near point when the person is wearing these glasses.

The Human Eye
(a)
(b)
do = ∞, di = −( 10.0 − 2.0)
= −8.0 cm
1
f
= 1
d o
+ 1
d i
= 1
∞ +
1
−8.0
( )
di = −( 8.0 − 2.0)
= −6.0 cm
1
d o
= 1
f
− 1
d i
=
1
( −8.0)
−
∴ N = 24.0 + 2.0 = 26.0 cm
= −
1
8.0 cm
We want a virtual,
upright image at the
person’s near point
We want a virtual,
upright image at the
person’s far point
⇒ f = −8.0 cm
f < 0, diverging lens
1
−6.0
( ) = 0.0417 cm−1 ⇒ d o = 24.0 cm
Close to the normal
near point of 25 cm

The Human Eye
FARSIGHTEDNESS
The lens creates an image of the close object at the near point
of the farsighted eye.

Example: Eyeglasses for the Farsighted Person
A farsighted person has a near point located at 75.0 cm. from the eye.
Assuming that eyeglasses are to be worn 2.0 cm in front of the eye,
find the focal length needed for the lens of the glasses so the person’s
near point with glasses is 25.0 cm.

do = 25.0 − 2.0 = 23.0 cm, di = −( 75.0 − 2.0)
= −73.0 cm
1
f
= 1
d o
+ 1
d i
f = 33.6 cm
= 1
23.0 +
1
−73.0
( )
f > 0, converging lens
= 0.0298 cm−1
We want a virtual,
upright image at the
person’s near point

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.
Refractive
power
(in
diopters)
=
f
1
( in meters)

Example: What refractive power, in diopters, would the optometrist
prescribe for the lenses found in the last two examples, i.e. nearsighted
and farsighted?
Refractive power (in diopters) = P =
Nearsighted example :
f = −8.0 cm = −0.080 m
P = 1
f =
1
−0.080
( )
Farsighted example :
f = 33.6 cm = 0.336 m
P = 1
f =
1
0.336
= −12.5 D
= +2.98 D
1
f in meters
( )

Angular Magnification and the Magnifying Glass
The size of the image on the retina determines how large
an object appears to be.

Angular Magnification and the Magnifying Glass
θ
( in radians)
=
Angular size
≈
h
d
o
o

Angular Magnification and the Magnifying Glass
Example: A Penny and the Moon
Compare the angular size of a penny held at arms length with that of
the moon.
Penny
Moon
θ ≈
θ
≈
h
d
h
d
o
o
o
o
1.
9 cm
=
71cm
3.
5×
10
=
3.9×
10
=
6
8
0.
027
m
m
=
rad
0.
0090
rad

Angular Magnification and the Magnifying Glass
Angular magnification
Special cases:
Relaxed focus
M
=
θʹ′
θ
di = −∞ ⇒ M ≈ N 1 %
'
& f
Focus at N
θ ' ≈ ho , θ ≈
do ho N
M ≈ h o d o
h o N
= N
d o
di = −N ⇒ M ≈ N 1 %
'
& f
= N 1 #
%
$ f
− 1
−∞
( )
− 1
−N
( )
(
*
)
= N
− 1
d i
(
*
)
= N
f
&
(
'
f +1

Example of a magnifying glass: A magnifying glass has a focal
length of 9.5 cm. Find the magnification of the lens when used
(a) with a relaxed eye, and (b) with the image at N=25 cm.
f = 9.5 cm
(a) M ≈ N
f
(b) M ≈ N
f
25
=
9.5 = 2.6 times d ( o = f = 9.5 cm)
25
+1=
9.5 +1= 3.6 times d ( o = 6.9 cm)