This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Practice Midterm 1 Solution: Please email Aritoki Suzuki (asuzuki@berkeley.edu) for questions. 1 Crystal Ball a) Where does the image form? Image formed by refraction off glass surface can be calculated by n 1 d o + n 2 d i = n 2 n 1 R (1) 1 3 r + 1 . 5 d i = 1 . 5 1 r (2) d i = 9 r (3) So image forms 7 r behind silvered mirror. Now we use mirror equation to find out where image get formed by mirror 1 d o + 1 d i = 1 f (4) 1 7 r + 1 d i = 1 r 2 (5) d i = 7 15 r (6) So image will be formed 23 15 r to right of front of glass sphere, thus we again calculate image formed by refraction at surface of lens n 2 d o + n 1 d i = n 2 n 1 R (7) 1 . 5 23 15 r + 1 d i = 1 . 5 1 r (8) d i = 23 11 r (9) So image forms 1 11 r behind silvered mirror! b) Find magnification. To find total magnification from refractions and mirror, we can find out 1 magnification for each step and multiply them. Rewriting equation we used for first refraction, 1 3 r + 1 . 5 d i = 1 . 5 1 r (10) 1 4 . 5 r + 1 d i = 1 3 r (11) We see that refraction has same effect as having lens with focal length 3 r and object distance of 4 . 5 r . Thus magnification is m 1 = d i d o = 9 r 4 . 5 r = 2 (12) For mirror, we can simply use magnification equation for mirror m 2 = d i d o = 7 15 r 7 r = 1 15 (13) For second refraction we again rewrite equation 1 . 5...
View
Full
Document
This note was uploaded on 10/05/2009 for the course PHYSICS 7C taught by Professor Lin during the Spring '08 term at University of California, Berkeley.
 Spring '08
 LIN
 Quantum Physics

Click to edit the document details