MT1 Review Problems Solutions - Practice Midterm 1...

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Unformatted text preview: Practice Midterm 1 Solution: Please e-mail 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...
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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.

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MT1 Review Problems Solutions - Practice Midterm 1...

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