Announcements 11/9/11 Prayer Term project progress report due Saturday

nighta. Evidence of progressb. Text in body of email, who is in your group,

CC group members Slinkies! (Thomas, Mike, Rachael, this means

you)

Frank & Ernest (with apologies

to Pluto)

Thought question Which will look bigger* to you, a 1 m tall

object that’s 5 meters away from you, or a 10 m tall image that’s 50 meters away from you?

a. 1 m tall objectb. 10 m tall imagec. same

* In the sense that it takes up more of your field of view

“angular size”

Worked Problem What is the angular size of a 0.1 m tall

object that’s 5 meters away from you?

Two methods!

Thought Question Which method should you use?

a. Method Ab. Method B

Quick writing

You are looking at an ant, h = 1 mm. What is the maximum viewing angle you can use to look at the ant, without any lenses?

r

“Colton picture”

(in radians) = (section of arc)/r

Reading Quiz Which of the following is NOT true of angular

magnification?a. It is more useful than the absolute magnification

when discussing telescopesb. It is more useful than the abs. magnification when

discussing magnifying glassesc. It is given by the equation m = -q/pd. It is likely to show up on an exam.

m = /0 … where 0 = “the best you can do without magnification”

Magnifying Glass The setup:

f = 10 cm

Where would you like the image to be? Let’s pick q = -50 cm. (q would generally be given

in problem.) What is m? (m = /0)

a. What is ?b. What is 0?

Answers: = 6h/50 rad 0 = h/25 radm = 3

Note: using formulas from book…mmax = 3.5 (for q = 25 cm)mmin = 2.5 (for q = infinity)

Aside: What will the ant see?

8.33 cm

h

p

f = 10 cm

If p = 1 cm: q = (1/10 – 1/1)-1 = -1.11

M = -q/p = 1.11

m = /0 = 1.11 9.44

1.109.33

h

h

If p = 9.5 cm: q = (1/10 – 1/9.5)-1 = -190

M = -q/p = 190/9.5 = 20

m = /0 = 20 (190 8.33)

1.80(9.5 8.33)

h

h

If p = 20 cm: q =+20

(ant can’t see you!)

(fixed from in-class values)

Quick writing

You are looking at the planet Mars, “h” (diameter, really) = 3.4 106 m. The planet, as you are looking at it, is 2.5 1011 m away (this changes from month to month based on the relative positions of Mars and Earth). What is the maximum viewing angle you can use to look at Mars, without any lenses?

r

“Colton picture”

(in radians) = (section of arc)/r

Telescope

The setup:

Given details of setup, what is m? (m = /0)

a. What is 0?

b. What is ?

These focal spots should essentially overlap (not shown properly in this figure)

“Colton picture” for

Because Mars is so far away, image is formed at the focal spot (essentially)

fo

Height of image = hfo/r (from M = -q/p)

r

fe

image

If intermediate image were formed exactly at the focal point of the eyepiece, final image would be at . As it is, it will just be very far away.

Regardless of how far away it is, though, the angle is given by the blue ray.

triangle: (rad) = (intermed. height)/fe

Answers:0 = h/r = foh/(rfe)m = fo/fe

Reflecting Telescope

A “Newtonian Reflector”

http://lcogt.net/en/book/reflecting-telescopes

Incoming Light

Mirror

CurvedMirror

eyepiece lens

Compound Microscope

http://en.wikipedia.org/wiki/Microscope

I really dislike the eqn: “overall magnification” =

Mobjective meyepiece

because it mixes absolute magnification with angular magnification

(but apparently everyone does it that way)

Not on reading assignment, not on HW, not on exam, not especially interesting… let’s not bother with. Onward!

Chapter 37!

Interference effectsa. I.e. now returning to wave nature of

light, instead of the ray approximation

Two mathematical facts we will use:

cos2

ix ixe ex

sin

2

ix ixe ex

i

Interference... Interference...

A single source

Next few slides: credit Dr. Durfee

Interference... Interference...

Two sources