announcements 11/2/12

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Announcements 11/2/12 Prayer Slinkies! (Seth, Ryan, William B, Clement) Progress Report due a week from Saturday Pearls Before Swine

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Announcements 11/2/12. Prayer Slinkies! (Seth, Ryan, William B, Clement) Progress Report due a week from Saturday. Pearls Before Swine. From warmup. Extra time on? (nothing in particular) Other comments? - PowerPoint PPT Presentation

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Page 1: Announcements 11/2/12

Announcements 11/2/12 Prayer Slinkies! (Seth, Ryan, William B, Clement) Progress Report due a week from Saturday

PearlsBeforeSwine

Page 2: Announcements 11/2/12

From warmup

Extra time on?a.(nothing in particular)

Other comments?a.Do astronomers have to worry about

the fact that air doesn't have an n of exactly 1?

Page 3: Announcements 11/2/12

Clicker question: A converging lens should be:

a. convexb. concave

Page 4: Announcements 11/2/12

Convex/converging Lens

parallel rays converge to the focal point

fif not parallel, they converge (image formed) at a different point

at each interface, Snell’s law is

satisfied

Page 5: Announcements 11/2/12

Where will image be formed?

focusfocus

The equation:

1 1 1

p q f

Ray diagram just like one for concave mirror, except rays continue to right instead of reflecting back!

positive q = right hand side of lens q

Mp

Magnification same:

Page 6: Announcements 11/2/12

From warmup

The mirror eqn and the thin lens eqn look identical. How exactly are they different?

a.The sign conventions are different, a positive focus means a focus that is "behind" the lens. The same is true of q.

b.(from my answer) Also--how f is calculated is different. For mirrors, it's simply related to the radius of curvature, but for lenses it relates to the curvature as well as the index of refraction of the lens glass.

Page 7: Announcements 11/2/12

Clicker quiz

Just out of curiosity… have I ever used your warmup answer?

a.yesb.no

Page 8: Announcements 11/2/12

Concave/diverging Lens

parallel rays seem to be coming from the focal point

f

Page 9: Announcements 11/2/12

Diverging lenses

virtual image!The equation:

1 1 1

p q f (negative f)

Negative q: means image on same side of lens as object (virtual)

Page 10: Announcements 11/2/12

Demos

Demo: “Real image or not?” Class poll: What kind of image is this?

a. Realb. Virtual

Page 11: Announcements 11/2/12

Sign conventions: summary Assuming that the light rays are coming from the

left:quantity positive

+negative

p if on the left of the

lens/mirrorif on the right of the lens/mirror

(can only happen in a compound problem)

q if on the right of the lens (left side of mirror), means a real image

if on the left of the lens (right side of mirror), means a virtual image

f if converging: convex lens/concave mirror

if diverging: concave lens/convex mirror

M if image is right-side up if image is upside-down

Page 12: Announcements 11/2/12

“Lensmaker’s Eqn”

1 2

1 1 11n

f R R

Lens-makers’ eqn:

Page 13: Announcements 11/2/12

From warmup When you look straight

down at an object under water, will it appear to be closer to you or farther away than actuality? Why?

a. Because you are looking at a flat refracting surface the image of the object forms closer to the surface than the actual object is (figure 36.18). This means the object appears closer to you than it actually is.

Page 14: Announcements 11/2/12

Clicker Quiz Is that a real or a

virtual image?a. realb. virtual

Page 15: Announcements 11/2/12

Thick lenses and surfaces

1 2 2 1n n n n

p q R

Careful with signs!!!

(see table 36.2)

Page 16: Announcements 11/2/12

Clicker question: Can a converging lens ever produce a

virtual image?a. Yesb. No

Page 17: Announcements 11/2/12

Clicker question: What happens if I cover up the upper half of

lens?a. Nothingb. Upper half of image disappearsc. Lower half of image disappearsd. Intensity of image goes down by 50%

Page 18: Announcements 11/2/12

Clicker question: Can a diverging lens ever produce a real

image?a. Yesb. No

Page 19: Announcements 11/2/12

Multiple element problems Image of one becomes object of next Worked problem: Find qfinal, Mtot

60 cm 15 cm

f = 20 cmf = -60 cm

Answers: image is 20 cm to right of -60cm lens; M = -0.67