Transcript f = l - UCSD Department of Physics

A camera.
Where is the film with respect to the focal point?
Imagine a camera with a single lens with a focal distance f = 35 mm.
By how much and in what direction should the lens be moved to
move its focus from an object, which is far-far away to one at a distance
of 1.5 m?
Far-far away means l = , and l’ = f = 35 mm.
1 1 1
 
l l' f
New distance l = 1.5 m; l’- ?
1 1
lf
1500 mm  35 mm
l '  1 /(  ) 

 35.8 mm
f l
l f
1465 mm
l  l  l ' 0.8 mm
A movie projector.
Where is the film with respect to the focal point?
The diameter of an eyeball is about 23 mm.
It focuses light emanating from different objects onto the retina to produce
sharp images.
Most of the focusing job is done by the cornea, which acts as a fixed lens
with a focal distance of 23 mm.
The lens is only doing some fine tuning to move the focusing from objects
which are far-far away to objects which are close.
A healthy human eye can
clearly see (focus) objects at
distances from infinity to about
25 cm. How is that achieved?
By changing the focal
distance of the lens! We
always have l’ 23 mm.
1 1 1
 
l l' f
Far-far away means l = , and f = l ’  23 mm.
An object at l = 25 cm means
1 1
ll '
250 mm  23 mm
f  1 /(  ) 

 21 mm
l' l
l  l'
273 mm
The adaptive lens driven by the eye muscles changes the focal
distance of the eye by “only” 10%. But this is quite a lot!..
Accommodation from infinity to
about 25 cm.
How much does the “strength”
of the lens change?
The strength of a lens is called
refractive power. It is
measured in diopters.
1
dpt 
f
f is in meters.
When the eye accommodates from infinity to 25, its refractive power
increases by
1 1
1
1
dtp 
f2

f1

0.021m

0.023 m
 4dpt
It is equivalent to putting in front of the eye a lens with refractive
power of 4 dpt and a focal length of
f '  1/ 4 dpt  0.25 m  25 cm
Myopic (shortsighted) eye – the lens is always “too strong”, that is
too much converging and if the object is far away it creates its
image in front of the retina.
So, the light pattern on the retina becomes blurry, out of focus.
The eye can be helped by a negative, diverging lens, which creates
virtual images of far away objects closer to the eye.
Rephrasing it: the diverging lens + the “too much” converging lens
in the eye make a composite lens of the right converging power.
Hyperopic (farsighted eye) – the lens is sometimes not quite strong
enough.
If an object is close it focuses behind the retina.
So, the light pattern on the retina becomes blurry, out of focus.
The eye can be helped by a positive, converging lens, which
creates magnified virtual images of close objects further away from
the eye.
Rephrasing it: the additional converging lens + the “not strong
enough” converging lens in the eye make a composite lens of the
right converging power.
Optical instruments: magnifying glasses, spyglasses, telescopes…
What do they do?
b
a
The mother elephant looks larger than the cub elephant, because the
mother subtends larger angle, b > a , and has large angular size.
Therefore, we can better see mother’s ear than cub’s ear. The situation
changes if the cub is closer to the observer.
a
A spyglass creates a virtual image, which is closer to the observer and
has a larger angular dimension. So, one can better see the details
Magnifying glass.
Your eye cannot see objects
clearly from closer than the
“near point”, L = 25 cm.
So, the largest angular size of
a fine-print object with height h
is a = h/L.
A magnifying glass creates a
virtual image with the same
angular size as the object, but
you can now have the object at
a small distance f from your
eye. So, the angular size is
b = h/f.
Magnification of the magnifying glass:
M  b / a  L / f  25cm / f
If you go along the
green line (upward),
you will see a
pattern of nodes and
antinodes like in a
standing wave.
Two sources oscillating in phase.
Lines of nodes and antinodes?
Are antinodes the points, where disturbance is always high?
Are nodes the point, where disturbance is always low?
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