# Let the location of the image be denoted as q q p h h

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Let the location of the image be denoted asq.orR=2p qp+q=2 (41.3 cm) (289.1 cm)41.3 cm + 289.1 cm=72.275 cm.01210.0pointsAsphericalChristmastreeornamentis3.76 cm in diameter.What is the magnification of the image ofan object placed 13.3 cm away from the orna-ment?Correct answer: 0.0660112.Explanation:Let :p= 13.3 cmandR=-3.76 cm2=-1.88 cm.qphh'fRSince the image is 7 times larger than theobject, and is a real image, as stated in theproblem (p >0 andq >0), the magnificationmust be negative;i.e.,M=-7.M=hh=-qp=-7,soq=-M p=-(-7) (41.3 cm) =289.1 cm.011(part2of2)10.0pointsWhat is the radius of curvature of the mirror?RorR=2p qp+q=2 (41.3 cm) (289.1 cm)41.3 cm + 289.1 cm=72.275 cm.01210.0pointsAsphericalChristmastreeornamentis3.76 cm in diameter.What is the magnification of the image ofan object placed 13.3 cm away from the orna-ment?Correct answer: 0.0660112.Explanation:Let :p= 13.3 cmandR=-3.76 cm2=-1.88 cm.
segovia (cs39966) – Homework 8 – Spurlock – (14441)5Using the mirror equation,1f=2R=1p+110.0pointsThe magnification produced by a converginglens is found to be 2.7 for an object placed21 cm from the lens.What is the focal length of the lens?01310.0pointsA convex mirror has a focal length of-24 cm.What is the object distance if the imagedistance is-17.1 cm?> p > ff < q <0> m >-∞f > p >0-∞< q <0> m >101410.0pointsThe magnification produced by a converginglens is found to be 2.7 for an object placed21 cm from the lens.What is the focal length of the lens?> p > ff < q <0> m >-∞f > p >0-∞< q <0> m >1What is the distance between the objectand the lens?
segovia (cs39966) – Homework 8 – Spurlock – (14441)61p+1q=1fM=hh=-qpDetermine the magnification of the image.016(part1of2)10.0points017(part2of2)10.0pointsDetermine the magnification of the image.
An object is located 28 cm from a diverginglens with a focal length of 36 cm.Determine the image distance.

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Term
Fall
Professor
HOFFMAN
Tags
Work, Snell s Law, Correct Answer, Total internal reflection, Segovia