Answers to 2016 Practice Exam IV (chapters 31‐35) Question 1. 2. 3. 4. 5. 6.NEW* 7. 8. Answer (a) m=1.25kg (b) k=372.2 N/m (c) X= 175 µm (d) v= 0.003 m/s C C D B L=0.025H i= 0.185A Gauss’ Law of electricity Page 949 summary. You must be able to do the derivation starting with Gauss. EQN 32‐11 on page 947 9. 10. 11. 12. 13. 14 15. 16. D B B B C E E 17. 18. 19. D E A 20. 21. B It is 16.67 cm to the right of lens two. It is real, non‐inverted, and the same size as the object. D Image is 12 cm to the left of lens, it is virtual, upright, and smaller (40% of its original size.) f= ‐ 60 cm 22. 23. 24. .
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25. Image is 2.8 cm to the left of the mirror, it is inverted, real, and smaller. 26. Lcommon=2.454 µm Comment if Necessary Ensure you look at NEW emailed #6. Delete the one in the packet. Look at page 666. Remember Gauss’ law has its own chapter too. Look at homework #32 if necessary for what terms mean. Look at drawing on page 997. It is actually the Law of Reflection! HOWEVER, the exponent should be to the zero power it is a typo! Definition of Diverge i>0 thus real M>0 thus non‐inverted M=1 thus same size Look at drawing (b) on page 1026 i<0 thus virtual m>0 thus non‐inverted m=.4 thus 40% of its original size R1 is ‐20 and R2 is +30, if you assume object is on left. If you assume object is on right then R1 is ‐30 and R2 is +20. Other way the f is negative 60! See me if you need help on the ray diagrams! i>0 thus real m<0 thus inverted m=.4 thus 40% of its original size This is one problem where I would use Excel if open book, open note exam! Answers to 2016 Practice Exam IV (chapters 31‐35) 27. 28. a. d=.57 mm . b. m=2 gives 672 nm m=3 gives 480 nm Second minimum is m=1! Smallest would be m=0, so Second smallest is m=1! Visible light is 430nm‐690nm. So only two fall in that range.