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Aberrations in Optical Components (lenses, mirrors)
1) Chromatic Aberrations: Because the index of refraction of the lens material depends
slightly on wavelength, the focal length also depends on wavelength, so different
wavelengths will form images at slightly different places.
(Doesn’t occur for mirrors, since  = ’ for all wavelengths)
2) Spherical Aberration (both lenses and mirrors):
We have made the paraxial approximation, that all rays
are near the principal axis. As rays get further from the
principal axis, their focus point changes, so if the lens or
mirror collects light through a large diameter aperture,
the image will be blurry.
To sharpen the image, use a smaller diameter aperture
in front of or behind the lens (or in front of the mirror)
to restrict the light to being close to the axis;
in a camera this requires a longer exposure
time to compensate for the reduced amount
of light entering the camera/second.
aperture
A converging lens forms a real image
on the film or CCD detector.
In more expensive cameras, the
distance between the lens and
CCD/film (q) is adjustable so that
1/q = 1/f - 1/p to keep the image
sharp. An expensive camera will also
have an adjustable aperture so one
can limit spherical aberrations,
compensating by increasing the
exposure time.
Camera
or film
Less expensive cameras have a fixed
lens position
i.e. (lens-aperture distance = f)
and aperture, with diameter D <<f.
The image will be in reasonable focus
on the CCD/film as long as the image
distance p >> f.
(i.e. 1/q = 1/f – 1/p  1/f if p >> f)
(In the limit, D 0, you don’t even
need a lens: the camera is a pin-hole
camera: only “one ray” from each point
on the object reaches the CCD/film.
These are sometimes used for
surveillance, because difficult to
detect.)
f
Cameras in cell phones typically replace the mechanical shutter with an “electronic
shutter” which turns the pixels in the CCD detector on/off; this can sacrifice resolution
of the pixels. Also, better cell phone cameras have an autofocus, that moves the lens
slightly to get better focus.
Eye
• Pupil is aperture which controls how much light enters eye.
• Real image formed on retina where light sensors (rods and cones)
send signals through optic nerve to brain.
• Refraction occurs in outer surface (tears, cornea), aqueous humor,
and lens.
• Ciliary muscles can cause the lens to contract,
decreasing its focal length, to keep image in focus
on retina.
For a “normal” eye, parallel
rays from  are focused on the
retina when the lens is
“relaxed” (i.e. not
compressed). Therefore, the
normal eye can focus on .
For rays from a closer object to be
focused on the retina, the lens must
be compressed to decrease f:
1/f = 1/p + 1/q, to keep q = size of
eyeball.
The shortest distance that the eye
can focus on = near point. For a
normal eye, near point  25 cm.
If the object is closer than the near
point, it will be focused outside the
eye; i.e. the image on the retina will
be blurry.
f
an
A farsighted person’s near point is larger than 25 cm. This can happen because the
eyeball is “too short” or because the lens cannot compress (and reduce its focal
length) enough. (The latter often happens with old age as the lens becomes less
flexible or the ciliary muscle weakens.)
A farsighted person’s image can be corrected by placing a converging lens in
front of the eye, which will bring the focus onto the retina.
One needs f(lens) > object distance, so that the lens forms a virtual image
at the near point of the eye when the object is 25 cm from the eye.
Consider the case where the lens is very close to the eye (e.g. contact lens) so that image
and object distances can be measured from the eye.
One wants the position of the virtual image of the correctional lens (q)
q = - patient’s near point when p = 25 cm (normal near point):
 1/f = 1/25cm - 1/near point.
Example, if near point = 50 cm, one needs 1/f = 1/25cm - 1/50cm = +1/50cm
f = 50 cm
For a nearsighted eye, the focal length of the relaxed eye-lens (i.e. its largest value) is
less than the length of the eyeball, so the eye cannot focus on objects at infinity (i.e.
the eyeball is “too long”.
The most distant object that can be brought to focus on the retina
(i.e. by the relaxed eye-lens) is at the “far point”.
far point
[For the normal eye, the far point = .]
an
The diverging lens should form a virtual image at the eye’s far point for an object at
“infinity”.
Object at 
Image at far point
-q
p
For example, if the diverging lens is very close to the eye (e.g. contact lens) so all
distances can be measured from the eye, 1/p + 1/q = 1/f
1/ - 1/far pt. = 1/f
f = - far pt.
diverging lens
Magnifying Glass
Recall that a converging lens will form a magnified virtual image if the object
is inside the focal point (p < f). Since 1/q = 1/f - 1/p  q = -pf/(f-p).
Therefore, the lateral magnification M = -q/p = f/(f-p) > 1.
It is more useful to consider the “angular magnification”, which compares how
large a viewer perceives the image to be with how the large the object could
appear without use of the lens; i.e. compares the sizes of the images (with and
without the magnifying glass) on the retina.
For an object a distance p from the eye:
The size of the image of an
object on the retina    h/p
(tan  ).
Since the minimum value of p for
which the eye can form a clear image
is the near point, the image on the
retina will have its maximum size
when the object is at the eye’s near
point:
maximum size (without magnifier) =
0  h/LNP
LNP
LL = -q
The angle for the image formed
by the magnifying glass is
 = h’/L = h/p
To see this image clearly, one
needs L to be between the
eye’s near point (LNP) and its
far point (LFP).
Since L = -q = pf/(f-p)  p = Lf /(L+f)
so LNPf/(LNP+f) < p < LFPf/(LFP+f)
Therefore, the angular magnification
m = /0 = (h/p) / (h/LNP) = LNP/p
(LNP/LFP) (LFP/f +1) < m < (LNP/f + 1)
image at far point: relaxed eye
image at near point:
maximum m
(LNP/LFP) (LFP/f +1) < m < (LNP/f + 1)
For normal eye, LNP = 25 cm and LFP =  
25 cm/f < m < 25cm/f + 1
relaxed eye
Example: If f = 6 cm,
If f = 2.5 cm
maximum m (image at NP = 25 cm)
4.167 < m < 5.167
10 < m < 11
The smaller f [i.e. the “fatter” the lens], the larger m.