Optical instruments
PHY232
Remco Zegers
zegers@nscl.msu.edu
Room W109 – cyclotron building
http://www.nscl.msu.edu/~zegers/phy232.html
quiz (extra credit)
¾ unpolarized light with intensity I0 is projected onto a
polarizer whose polarization axis is at 450 relative to the
vertical axis. The light that passes through is then
projected onto a second polarizer with a polarization axis
that is at 900 relative to the vertical axis. Which of the
following is true?
¾ a) the intensity after the first polarizer is the same as I0
¾ b) the intensity after the second polarizer is 0
¾ c) the intensity after the second polarizer is smaller than I0,
but larger than 0.
¾ d) the intensity of the light after the second polarizer is
larger than the intensity after the first polarizer
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optical instruments
¾ Most optical instruments involve just the laws of reflection
and refraction – microscope, telescope etc
¾ some optical instruments make use of the wave-nature of
light, such as the interferometer
¾ In this chapter we consider some optical instruments,
starting with the eye
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the eye
¾ the eye essentially consists of a lens that focuses light on
the retina. The ciliary muscles are used to change the
curvature of the lens and hence the focal length.
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the eye II
¾ when the ciliary muscles are relaxed, an object at infinity is focused
onto the retina. The focal length is about 1.7 cm.
¾ optometrists define the power P of a lens in terms of diopters
¾ D=1/f (f in m, D in diopters 1/m)
¾ the typical eye has a power of 1/0.017 m=59 diopters
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the far-point
¾ The largest distance that can clearly be seen is called the far-point FP.
¾ a good human eye can visualize objects that are extremely far away
(moon/stars) and the far point is then close to infinity.
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nearsightedness (myopia)
¾ In case of nearsightedness, the
far-point is much smaller than
infinity for example because
the eyeball is elongated.
¾ on object placed at infinity is
focused in front of the retina.
¾ this can be corrected using a
diverging lens…
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example
¾ A person cannot see objects clearly that are more than 50
cm away from his eye. An optometrist therefore prescribes
glasses to solve the problem. What should the power of
the lens be (in diopters) to solve the problem? You can
ignore the distance between the glasses and the eye lens.
without
glasses
answer: to solve the problem,
the lens should be made such
that the image of an object at
infinity is projected at the far-point.
1/p + 1/q = 1/f
1/∞ + 1/(-0.5) = 1/f (virtual image)
f=-0.5 so D=1/-0.5=-2.
-
-
-
20/20 vision: you can see on the chart what average people
can see from 20 feet
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the near point
¾ The near-point is the closest distance in front the eye that
a person is capable of focusing light on the retina
¾ the near-point for a normal person is about 25 cm, making
it hard to focus an object closer to you eyes than that.
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farsightedness (hyperopia)
¾ happens when the nearpoint is much larger than
25 cm.
¾ it becomes hard to see
objects nearby since the
eye muscles cannot
accommodate it.
¾ it can be corrected using a
converging lens (reading
glasses)
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example
¾ a person suffering from hyperopia has a near point of 1 m.
The optometrist has to prescribe a lens of what power to put
the near point back at 25 cm?
To solve this problem, you have to realize that
the lens must be constructed such that an
object situated at the desired nearpoint (25 cm)
is imaged onto the nearpoint of the person.
p=0.25 m
q=-1 m
1/p + 1/q = 1/f
1/0.25 + 1/(-1)=1/f
1/f=P=3 diopters
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question
¾ a person has a far-point of 1 m and a near point of 75 cm.
In order to help this person see objects that are far away
and allows him/her to read a book…
¾ a) bi-focal glasses are needed, which are partly diverging
and partly converging
¾ b) glasses are needed that bring the far-point and the
near-point together
¾ c) an operation is needed to solve at least one of the
problems so that the other can be solved with glasses
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lon-capa
¾ do questions 10,11 from set 9
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simple magnifier
θ0
h
eye
NP
h’
h
p
f
q
the best result is obtained
if the image is at infinity
(eye relaxed). to do so p=f
and m=NP/f
PHY232 - Remco Zegers
¾ normal eye cannot focus if the object
distance < near point (NP)
¾ therefore, the maximum subtended angle
equals:
θ0=h/NP=h/(25 cm)
(assumed that tanθ=θ θ: small
¾ if we put the same object in front of a
converging lens with p<f, a virtual upright
image is created
¾ 1/p+1/q=1/f with p<f
¾ q=pf/(p-f) with p<f so q: negative
¾ M=himage/hobject
=-q/p=-f/|f-p| with p<f so M>1
¾ maximum subtended angle now equals:
θI=h’/q=h/p
¾ angular magnification m=θI/θ0
¾ m=θI/θ0=(h/p)/(h/NP)=NP/p
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example
¾ A lens with f=10 cm is used as a magnifier. What is the angular
magnification if the image if formed at the near point?
¾ What is the angular magnification if the eye is relaxed (image at
infinity?)
a) q=-25 cm (near point, but virtual)
m=NP/p=-q/p=M (angular magnification: lateral
magnification)
1/p + 1/q = 1/f 1/p + 1/-25 = 1/10
p=7.14 cm
M=m=-q/p=-(-25)/7.14=3.5
b) q=∞, so p=f and m=NP/f=25/10=2.5
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the microscope…
L
…uses two converging lenses with focal lengths f1 and f2 with f2>f1
moreover, L>>f2
For lens 1: p1 is chosen such that q1~L (image 1 will appear just within F2)
This happens when p1~f1
so: M1=-q1/p1=-L/f1
Lens two then acts as a magnifying glass with m2=NP/f2=25/f2
The magnifying power is defined as m=M1m2=(-L/f1)(25/f2)=-25L/(f1f2)
(all units in cm), usually written as: m=-25L/(fOfe) (inverted!)
with O for objective and e for eyepiece
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example
A red blood cell has a size of about 7x10-6 m. A
microscope is used to visualize it. The microscope
has L=30cm, f0=1 cm, fe=0.5 cm. How large is the cell
when seen through the microscope?
answer: m=-25L/(fOfe)=-25x30/(1x0.5)=-1500
virtual size=real size x m=7x10-6 (m)x –1500=-0.01 m
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L
a telescope
size of image of
objective lens: hi
θe
θo
¾ a telescope is very similar to a microscope except that the lenses are
slightly differently configured:
¾ light comes in (from a star) almost parallel. It is focused at the focus
point fo of the first converging (objective) lens.
¾ this image becomes the object for the second converging (eyepiece)
lens and is place just at the focal length fe of that lens.
¾ tan(θo)≈θo=hi/fo
¾ tan(θe)≈θe=hi/fe
¾ magnifying power m= θe/θo=fo/fe note L=f0+fe
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example
¾ A telescope has two lenses which are 92 cm apart from
eachother. The angular magnification of the telescope is
45. What are the focal lengths of the objective and eyepiece lens?
m= θe/θo=45=fo/fe
L=f0+fe=92
combine: 45=(92-fe)/fe
solve for fe=2.1 cm
find fo=89.9 cm
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loncapa
¾ do problem 12 from lon-capa 9
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resolution
¾ resolution: the ability of an optical system to distinguish
between two closely spaced objects
¾ resolution is limited by the wave nature of light: when light
passes through a slit, it is diffracted and thus smeared out.
¾ if the angular separation becomes two small, objects
become hard to distinguish
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rayleigh’s criterion
¾ two images are just resolved if rayleigh’s criterion is fulfilled.
¾ Rayleigh’s criterion: the central maximum of image A false into the
first minimum of image B
¾ first diffraction minimum: sinθ≈θ=λ/a with A the slitwidth
¾ images separated by a minimum angle θmin=λ/a can just be resolved
¾ if the aperture is circular with diameter D: θmin=1.22λ/D
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¾θmin=1.22λ/D
question
¾ a binary star system consists of two stars that are rotating
around each other. Because of there closeness they are
hard to separate. A color filter can be used to improve the
separation. Is it better to use blue or red to make a picture
that best separates the stars?
a) Blue
b) Red
c) doesn’t matter
if λ is low, θmin low: one can separate images that are closer together
better with blue light than with red light
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lon-capa
¾ do problem 13 from lon-capa 9
¾
¾
¾
¾
use θmin=1.22λ/D
distance earth moon is 3.84E+8 m
how to calculate θmin? you can use tanθ=θ (radians)
use a wavelength of 550 nm
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The Michelson Interferometer
moveable
1)
2)
3)
4)
5)
6)
monochromatic light is incident on mirror
light travels to moveable mirror and is… 1
reflected
some light is also passed through and is …
reflected
beam 3) and 5) interfere and make an
interference pattern
7) by moving the moveable mirror, the path
length difference can be varied
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2
5
6
4
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Michelson interferometer
moveable
3
2
5
1
6
4
¾ the compensator servers to make sure that the light going to either
branch travels the same distance through the glass.
¾ The path length difference D=2d23-2d45
¾ If the movable mirror moves by λ/2, D changes by λ and the
interference pattern is shifted by 1 fringe.
¾ Insertion of a material with index of refraction n in path 2-3 will also
make a path length difference, and by observing the change in the
interference pattern, one could determine n
¾ more about this in the last week’s lecture on relativity…
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