optical instruments phy232 remco zegers [email protected] room w109 – cyclotron building...

Optical instruments

PHY232Remco [email protected] W109 – cyclotron buildinghttp://www.nscl.msu.edu/~zegers/phy232.html

PHY232 - Remco Zegers - optical instruments 2

quiz (extra credit)

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?

a) the intensity after the first polarizer is the same as I0

b) the intensity after the second polarizer is 0 c) the intensity after the second polarizer is smaller

than I0, but larger than 0. d) the intensity of the light after the second polarizer

is larger than the intensity after the first polarizer


optical instruments

Most optical instruments involve just the laws of reflection and refraction – microscope, telescope etc

some optical instruments make use of the wave-nature of light, such as the interferometer

In this chapter we consider some optical instruments, starting with the eye


the eye

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.


the eye II

when the ciliary muscles are relaxed, an object at infinity is focused onto the retina. The focal length is about 1.7 cm.

optometrists define the power P of a lens in terms of diopters D=1/f (f in m, D in diopters 1/m) the typical eye has a power of 1/0.017 m=59 diopters


the far-point

The largest distance that can clearly be seen is called the far-point FP.

a good human eye can visualize objects that are extremely far away (moon/stars) and the far point is then close to infinity.


nearsightedness (myopia)

In case of nearsightedness, the far-point is much smaller than infinity for example because the eyeball is elongated.

on object placed at infinity is focused in front of the retina.

this can be corrected using a diverging lens…


example 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.

answer: to solve the problem,the lens should be made such that the image of an object atinfinity is projected at the far-point.

1/p + 1/q = 1/f1/ + 1/(-0.5) = 1/f (virtual image)f=-0.5 so D=1/-0.5=-2. - - -

withoutglasses

20/20 vision: you can see on the chart what average peoplecan see from 20 feet


the near point

The near-point is the closest distance in front the eye that a person is capable of focusing light on the retina

the near-point for a normal person is about 25 cm, making it hard to focus an object closer to you eyes than that.


farsightedness (hyperopia)

happens when the near-point is much larger than 25 cm.

it becomes hard to see objects nearby since the eye muscles cannot accommodate it.

it can be corrected using a converging lens (reading glasses)


example

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 mq=-1 m1/p + 1/q = 1/f1/0.25 + 1/(-1)=1/f1/f=P=3 diopters


question

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…

a) bi-focal glasses are needed, which are partly diverging and partly converging

b) glasses are needed that bring the far-point and the near-point together

c) an operation is needed to solve at least one of the problems so that the other can be solved with glasses


lon-capa

do questions 10,11 from set 9


simple magnifier normal eye cannot focus if the

object distance < near point (NP) therefore, the maximum subtended

angle equals:

0=h/NP=h/(25 cm)

(assumed that tan= : small if we put the same object in front of

a converging lens with p<f, a virtual upright image is created

1/p+1/q=1/f with p<f q=pf/(p-f) with p<f so q: negative M=himage/hobject

=-q/p=-f/|f-p| with p<f so M>1 maximum subtended angle now

equals: I=h’/q=h/p

angular magnification m=I/0

m=I/0=(h/p)/(h/NP)=NP/p

eye

NP

0h

h’h

p

q

f

the best result is obtainedif the image is at infinity (eye relaxed). to do so p=fand m=NP/f


example

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?

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 = 1/10p=7.14 cmM=m=-q/p=-(-25)/7.14=3.5b) q=, so p=f and m=NP/f=25/10=2.5


the microscope…

…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

L


example

A red blood cell has a size of about 7x10-6 m. A microscope is used to visualize it. The microscopehas L=30cm, f0=1 cm, fe=0.5 cm. How large is the cellwhen 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


a telescope

a telescope is very similar to a microscope except that the lenses are slightly differently configured:

light comes in (from a star) almost parallel. It is focused at the focus point fo of the first converging (objective) lens.

this image becomes the object for the second converging (eyepiece) lens and is place just at the focal length fe of that lens.

tan(o)o=hi/fo

tan(e)e=hi/fe

magnifying power m= e/o=fo/fe note L=f0+fe

o

size of image of objective lens: hi

e

L


example

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 eye-piece lens?

m= e/o=45=fo/fe

L=f0+fe=92

combine: 45=(92-fe)/fe

solve for fe=2.1 cm

find fo=89.9 cm


loncapa

do problem 12 from lon-capa 9


resolution

resolution: the ability of an optical system to distinguish between two closely spaced objects

resolution is limited by the wave nature of light: when light passes through a slit, it is diffracted and thus smeared out.

if the angular separation becomes two small, objects become hard to distinguish


rayleigh’s criterion

two images are just resolved if rayleigh’s criterion is fulfilled. Rayleigh’s criterion: the central maximum of image A false into

the first minimum of image B first diffraction minimum: sin=/a with A the slitwidth images separated by a minimum angle min=/a can just be

resolved if the aperture is circular with diameter D: min=1.22/D


question

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?

min=1.22/D

a) Blueb) Redc) doesn’t matter

if is low, min low: one can separate images that are closer together better with blue light than with red light


lon-capa

do problem 13 from lon-capa 9

use min=1.22/D

distance earth moon is 3.84E+8 m how to calculate min? you can use tan= (radians)

use a wavelength of 550 nm


The Michelson Interferometer

moveable

1

2 3

4

5

6

1) monochromatic light is incident on mirror

2) light travels to moveable mirror and is…

3) reflected4) some light is also passed through

and is …5) reflected6) beam 3) and 5) interfere and make

an interference pattern7) by moving the moveable mirror, the

path length difference can be varied


Michelson interferometermoveable

1

2 3

4

5

6

the compensator servers to make sure that the light going to either branch travels the same distance through the glass.

The path length difference D=2d23-2d45

If the movable mirror moves by /2, D changes by and the interference pattern is shifted by 1 fringe.

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

more about this in the last week’s lecture on relativity…

optical instruments phy232 remco zegers [email protected] room w109 – cyclotron building...

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