Chapter 17 - Light and Image Formation
Chapter 17 - Light and Image Formation
Chapter 17 - Light and Image Formation
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsScenarioA mirror has a focal length of +10.0 cm.QuestionLocate <strong>and</strong> describe the image for an object distance of 25.0 cm.AnswerBecause the focal length of the mirror is positive, it is a concave mirrorWe expect the possibilities of both real real <strong>and</strong> virtual imagesBecause the object distance is larger than the focal length, we expect the image to be realTo find the image distance, we use 1/q = 1/f − 1/p <strong>and</strong> solve for qq = pfp − f(25.0 cm)(10.0 cm)= = 16.7 cm.25.0 cm − 10.0 cmThe magnification of the image isM = − q p= −16.7cm25.0 cm = −0.667.<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
Change language