Announcements 11/10/10 Prayer Still missing slinkies from: a. a.Stephen Grant b. b.Jon Mortensen c. c.Chris Read d. d.Roger Brown e. e.Tess Larson (She dropped the class; she’s just on the list here so I don’t forget about her)

Another compound lens problem Class designed Ray diagram = ? q = ? Overall magnification = ?

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. a.1 m tall object b. b.10 m tall image c. c.same

Reading Quiz Which of the following is NOT true of angular magnification? a. a.It is more useful than the absolute magnification when discussing telescopes b. b.It is more useful than the abs. magnification when discussing magnifying glasses c. c.It is given by the equation m = -q/p d. d.The effective distance of the unmagnified image from the eye is 25 cm for a magnifying glass and nearly infinite for a telescope e. e.It is likely to show up on an exam. m =  /  0 … where  0 = “the best you can do without magnification”

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

Magnifying Glass The setup: f = 10 cm Where would you like the image to be? Let’s pick q = 50 cm. (This would generally be given in problem.) What is m? (m =  /  0 ) a. a.What is  ? b. b.What is  0 ? Answers:  = 6h/50 rad  0 = h/25 rad m = 3 Note: using formulas from book… m max = 3.5 (for q = 25 cm) m min = 2.5 (for q = infinity)

Quick writing You are looking at the planet Mars, “h” (diameter, really) = 3.4  10 6 m. The planet, as you are looking at it, is 2.5  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. a.What is  0 ? b. b.What is  ? These focal spots should essentially overlap (not shown properly in this figure) “Colton picture” for  Answers:  = f o h/(rf e )  0 = h/r m = f o /f e

“Colton picture” for  Because Mars is so far away, image is formed at the focal spot (essentially) fofo Height of image = hf o /r (from M = -q/p) r fefe 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)/f e

Compound Microscope I really disliked the book’s somewhat arbitrary “overall magnification” = M objective  m eyepiece (but apparently everyone does it that way) M objective  -L/f o (ray diagram, object near focus of objective, image near eyepiece) m eyepiece between m max = cm/f e, and m min = 25cm/f e Not on HW, not on exam, not especially interesting…let’s not bother with any more.

Chapter 37! Interference effects a. a.I.e. now returning to wave nature of light, instead of the ray approximation Two mathematical facts we will use: