7. Optical instruments
1) Cameras
1 1
1


di do f
di  f
d0  f
There is a certain range of distances over which
objects will be in focus; this is called the depth of field
of the lens. Objects closer or farther will be blurred.
The amount of light that
enters the camera:
D
t  
 f 
2
di
f
m 
do
do
f
f  number 
D
D  lens opening
1
1a) Basic parts of a camera. Camera adjustments.
•Focusing: adjusting the position of the lens so that the image is positioned on the film
•Shutter speed: controls the amount of time light enters the camera
•f-stop: controls the maximum opening of the shutter. This allows the right amount of
light to enter to properly expose the film, and must be adjusted for external
light conditions.
A digital camera uses CCD sensors
instead of film. The digitized image
is sent to a processor for storage
and later retrieval.
2
Example: Suppose that a correct exposure is 1/250 s at f/11. Under the same
conditions, what exposure time would be needed for a pinhole camera if the
pinhole diameter is 1.0 mm and the film is 7.0 cm from the hole? The pinhole
camera uses a tiny pinhole instead of a lens.
t  1/ 250 s
f / D  11
D'  1.0 m m
f '  7.0 cm
t ' ?
2
2
 f ' D'
D
 D' 

t    t '    t '  t 
 f 
 f '
 f D
2
2
 70m m/ 1m m 
t '  1 / 250s
  0.16s
11


Example: What is the focal length of the eye-lens system when viewing an object
(a) at infinity, (b) 33 cm from the eye? Assume that the lens-retina distance is 2.0 cm.
d i  2cm
a)d 0  
b)d 0  33cm
f ?
1 1 1
 
do di f
a) f  di  2cm
b)
1
1
1


 f  1.9cm
f 33cm 2cm
3
2) The Human Eye
The human eye resembles a camera in its basic functioning,
with an adjustable lens, the iris, and the retina.
•Most of the refraction is done at
the surface of the cornea.
•The lens makes small adjustments
to focus at different distances.
4
3) Corrective Lenses
a) Nearsightedness can be corrected with a diverging lens
b) And farsightedness with a diverging lens
5
Far point: Farthest distance at which object can be seen clearly.
Normal FP is at infinity.
Nearsightedness: far point is too close.
Near point: Closest distance at which eye can focus clearly.
Normal NP is about 25 cm.
dg
Farsightedness: Near point is too far away.
To calculate lens power
we use lens equation:
d fp
1
1
1

 P
di do
f
dg
Correction for the distance
between glass and eye:
do  d p  d g  0
d i  d p  d g   0
d np
6
Example: A person cannot see clearly objects more than 70.0 cm away.
What power of lens should be prescribed if the glass is to be worn 1.00 cm in
front of the eye?
do  
d i  f  d fpd g 
d fp 70.0cm
P  1/ f
d g  1.00cm
di  0
P  1/d fpd g   1/ 0.69m  1.45D
P?
Example: A nearsighted person wears glasses whose lenses have power of -0.15D.
What is the person's far point if the glasses are very close to the eyes?
P  0.15D
d ge  0
f ?
1
1
f  
m  6.7m
P  0.15
f  6.7m
7
Example: What power lens is needed to correct for farsightedness where
the uncorrected near point is 75 cm?
d o  25cm
d i  75cm
P?
1
1
1

 P
di do
f
1
1
1 3
2
8
P




 2.7 D
 0.75m 0.25m 0.75m 0.75m 3m
Example: A farsighted person can read a newspaper held 25 cm from his eyes,
if he wears glasses of +3.33 diopters. What is this person's near point?
d o  25cm
P  3.33D
di  ?
1
1
1

 P
di do
f
1
1

 3.33D
d i 0.25m
1
1
10  12  10
2




di
0.25m 3m
3m
3m
d i  1.5m
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Vision underwater
Vision is blurry underwater
because light rays are bent
much less than they would be
if entering the eye from air.
This can be avoided
by wearing goggles.
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4) Aberrations of Lenses and Mirrors
a) Spherical aberration: rays far from the lens axis do not focus at the focal point
Solutions:
compound-lens systems;
use only central part of lens
b) Distortion: caused by variation in magnification with distance from the lens axis.
Barrel and pincushion distortion:
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c) Chromatic aberration: light of different wavelengths has different indices of
refraction and focuses at different points
Solution: Achromatic doublet, made of lenses of two different materials
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5) Magnifying glass (Simple magnifier)
A converging lens that allows one to focus on objects closer than the near point,
so that it makes a larger, and therefore clearer, image on the retina.
0
0
ho
 '
d0

h0
25cm
The power of a magnifying glass is described by its angular magnification:
 ' 25cm
M  

d0
25cm
if d i    d 0  f 
M
f
h
1 1
1
25cm
if d i  25cm 
 
 M
1  i
d 0 f 25cm
f
ho
Note!
do  f  di  0
di
d o  f  d i    m  

do
25cm
25cm
M 
1
f
f
12
Example: A magnifying glass with a focal length of 8.5 cm is used to read
print placed at a distance 7.5 cm. Calculate (a) the position of the image;
(b) the angular magnification
f  8.5cm
1
1
1
1



d i 8.5cm 7.5cm
63.75cm
d o  7.5cm
di  ?
M ?
d i  64cm
M
 ' 25cm 25cm



d0
7.5cm
M  3 .3
Example: A person uses a converging lens of focal length 5.0 cm as
a magnifying glass. What is the maximum possible magnification?
f  5.0cm
M max  ?
M 
 ' 25cm


d0
1
1 1
 
do
f di
d i  25cm
 1

 do

1
1
  
 max f 25cm
M max 
25cm
25
1 
1  6
f
5
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6)Telescopes
A refracting telescope consists of two lenses at opposite ends of a long tube.
The objective lens is closest to the object, and the eyepiece is closest to the eye.
fo  fe  l
The magnification is given by:

h / fe 
fo
'
M  

h / f o  f e

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Example: A student constructs an astronomical telescope with a magnification of 10.
If the telescope has a converging lens of focal length 50 cm, what is the focal length
of the eyepiece? What is the resulting length of the telescope?
M  10
f o  50cm
fe  ?
l ?
fo
'
M  

fe
f 0 50cm
fe 

 5.0cm
M
10
l  f o  f e  50cm  5cm  55cm
Example: A170x astronomical telescope is adjusted for a relaxed eye when
the two lenses are 1.25 m apart. What is the focal length of each lens?
M  170
l  1.25m
fo  ?
fe  ?
fo l  fe
l
M 

 fe 
;
fe
fe
M 1
fo  M fe ;
125 cm
fe 
 0.73cm
170  1
f o  170 0.73cm  124cm
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6a) Different types of telescopes
Astronomical telescopes need to
gather as much light as possible,
meaning that the objective must be
as large as possible.
Hence, mirrors are used instead of
lenses, as they can be made much
larger and with more precision.
A terrestrial telescope, used for viewing objects on Earth, should produce
an upright image. Here are two models, a Galilean type and a spyglass:
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7) Compound Microscope
A compound microscope also has an objective and an eyepiece; it is different from
a telescope in that the object is placed very close to the eyepiece.
The magnification is given by:
 25cm  d i 
 25cm  l  f e 
    


M  M e m0  
 f e  d 0 
 f e  d 0 
25cm  l
M 
f e f 0
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Example: A microscope has an eyepiece with a focal length of 3.5 cm. An object is
placed 1.5 cm in front of the objective lens and its image appears 21.0 cm behind
the objective lens. The microscope is adjusted for a relaxed normal eye. What is the
overall magnification of the microscope after the adjustment?
 25cm  d i 
  
M  M e m0  
 f e  d 0 
 25cm  21cm 
M 
 
  100
 3.5cm  1.5cm 
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