Ch 27 Optics and the Eye
Light passes through the cornea of the human eye
and is focused by the lens on the retina. The
ciliary muscles change the shape of the lens.
The eye produces a real, inverted image on the
retina. Why don’t things look upside down to us?
Near Point  closest point to the eye that the
lens is able to focus ≈ 25 cm from the eye
(changes with age)
Far Point  farthest point at which the eye can
focus; should be infinitely far away
The simplest camera consists of a lens and film in a dark box
The camera lens cannot change shape; it moves closer to or farther
away from the film in order to focus.
The f-number characterizes the size of the aperture:
The combination of f-number and shutter speed determines the
amount of light that reaches the film.
Lenses in Combination and Corrective Optics
In a two-lens system, the image produced by the
first lens serves as the object for the second
lens.
To find the image formed by a combination of
lenses, consider each lens in turn, starting with
the one closest to the object.
The total magnification is the product of the
magnifications of each lens.
A nearsighted person has a far point that is a
finite distance away, due to the lens focusing too
much; objects farther away will appear blurry.
Glasses are essentially a diverging lens that forms
an image of a distant object at the far point.
The strength of corrective lenses is usually
quoted as refractive power, which is the inverse
of the focal length measured in 1/m or diopters.
Refractive Power = 1 / f
Example
(a) near sighted people like to see clearly in
The relaxed eyes of a patient
have a refractive power of 48.5
diopters.
(a) Is this patient nearsighted
or farsighted?
(b) Since this patient is
nearsighted, find the far point.
(Treat the eye as a single-lens system,
with the retina 2.40 cm from the lens.)
the distance, thus calculate refractive power
for very distant objects.
1/f = 1 / di + 1/do
1/f = 1/0.024 + 1/∞
1/f = +41.7 diopters
Since this is less then the given refractive
power, we know this person is
sighted.
near
Farsighted --> cannot focus on close objects
– the near point is too far away.
The lens is not focusing enough
A converging lens further focuses the image and
moves the image past the Near Point.
Example
A farsighted person uses glasses with a refractive power of 3.4
diopters. The glasses are worn 2.5 cm from his eyes. What is this
person’s near point, when not wearing glasses? (b) What is
required refractive power for contacts?
Key Assumption:
1/f = 1/di + 1 /
do
(b)
(b)
1/f = 1 / di + 1 /do
48.5 = 1/0.024 + 1/ do
do = 14.6 cm
The near point is generally
corrected to be at 25 cm
from the eye (not lens).
3.4 = 1/di + 1/(.25-.025)
di = -0.957 meters (from the lens)
N = 95.7cm + 2.5 cm
N = 98.2 cm
1/f = 1 / di + 1 / do
1/f = -1/.982 + 1/(.25)
1/f = +2.98 diopters
The Magnifying Glass
A magnifying glass (convex lens) brings the near point closer to the eye.
A magnifying glass with focal length 20 cm is placed
15 cm to the right of an arrow. Where the image?
1 / f = 1 / do
1/20 = 1 / 15
di
= -60 cm
+ 1 / di
+ 1 / di
Object at Focal Point
tan θ = ho / N
θ = ho / N
(when θ is small; tan θ ≈ θ)
Now, place a converging lens whose focal
length is less than N very close to the
eye, and place the object at the focal
point of the lens. (f < N)
1 / f = 1 / do + 1 / di
1 / f = 1 / f + 1 / di
di
= infinity
Image at Near Point
In this case the magnification is
maximized if image is at the near
point
This gives the object a larger
angular size.
tan θ’ = ho / f
θ’ = ho / f
M=
hi / ho
= θ’ / θ
M = ho / f / h o / N
M=N/f
1 / f = 1 / do
+ 1 / di
1 / f = 1 / do
- 1/N
(neg because same side as object)
do
=fN/(N+f)
M ≈ di /
do
M ≈ N / [f N / ( N + f )]
M=1+N/f
The Microscope
mobjective lens = hi / ho
mobjective lens ≈ - di / do = - do / fobjective
If object is placed near the
focal point of objective lens
then
meyepiece = N / feyepiece
(this eq derived above “Object at Focal Point”)
The image of the objective
lens is formed at the focal
point of the eyepiece.
…which, in turn, forms the
eyepiece’s image at infinity.
So magnification of the microscope is
m = (mobjective lens) (meyepiece )
m = (-do/ fobjective) (N / feyepiece)
Telescopes
M = hi
=
θ’
M = ho / f
/
/
eye
ho
θ
/ ho / fobj
Object is at infinity
And need largest Objective lens possible (to collect light)
M = fobjective lens / feyepiece
WHY???
Example with Objective lens
1/f = 1/∞ + 1 / dimage
and the image forms at the focal
point…thus
f = dimage
Example
A Cassegrain astronomical telescope uses two
mirrors to form the image. The larger (concave)
objective mirror has a focal length of 50.0 cm. A
small convex secondary mirror is mounted 43.0 cm
in front of the primary. Light is reflected from
the secondary through a hole in the center of the
primary, thereby forming a real image 8.00 cm
behind the primary mirror. What is the radius of
curvature of the secondary mirror?
Since rays are originating from infinity, the image appears
at the focal length or at 50 cm, which is 7 cm beyond the
2nd mirror. So object for 2nd mirror is 7cm behind the 2nd
mirror.
1 / f2 = 1/ do + 1 / di
1 / f2 = 1/-7 + 1 /(43+8)
f2 = -8.11 cm
Mirrors: R = -2f
R = -2(-8.11)
R = 16.2 cm
Example
We have a lighted arrow at 22.5
cm and a convex lens (focal
length is 5 cm) at 12.5 cm and a
mirror at 0 cm. Where is the
final image formed and what is
the magnification?
1/f = 1/o + 1/i
1/5 = 1/10 + 1/i
2/10 = 1/10 + 1/i
1/10 = 1/i
i = 10 cm
which is at 2.5 cm from the
mirror
Now we perform a ray trace.
Place a virtual person some
where behind the image (which is
the object for the mirror.)
And using law of reflection
(incident angle = exit angle)
Look at the red (from object) and
green lines (from person). These
are equal from the normal formed
at the surface of the mirror.
The image from the mirror is located 2.5 cm behind the mirror.
This is the object for the last image through the lens.
1/f = 1
/ o
+ 1/i
1/5 = 1/(12.5 - -2.5) + 1/i
1/5 = 1/15 + 1/i
i = 7.5 cm
which yields a final position of 20 cm from mirror.
as always a positive image distance means the image is on the
OPPOSITE side of the lens.
Lens Aberrations
Spherical aberration occurs when light
striking the lens far from the axis does not
focus properly. It can be fixed by grinding
the lens to a precision, non-spherical shape.
Chromatic aberration occurs when different
colors of light focus at different points.
Chromatic aberration can be improved by combining
two or more lenses that tend to cancel each other’s
aberrations. This only works perfectly for a single
wavelength