P 2 a q 1 20 cm 30 cm 10 cm where distances are now

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p2=a-q1= (20 cm)-(30 cm) =-10 cm,where distances are now measured from thesecond lens, whose focal length isf2= 25 cm :1p2+1q2=1f2.This yieldsq2=p2f2017(part 2 of 3) 10 pointsWhat is the magnification of the second lens?Explanation:The magnification of the first lens is givenbyM1=-q1p1=-30 cm15 cm=-2.The total magnificationMof the two lensesis the product ofM1andM2:M=M1M2= (-2) (0.714286)=-1.42857.Hence, the final image is real, inverted, andenlarged.keywords:019(part 1 of 1) 10 pointsAn object 18 cm high is placed 10 cm infront of a convex mirror with a focal length of-5.8 cm.What is the image height?
Drews, Scott – Homework 12 – Due: Nov 26 2006, midnight – Inst: Ditmire9The magnification is given byM=-io.Hence, the image height is given byH=h M= (18 cm)--3.67089 cm10 cm=6.6076 cm.keywords:020(part 1 of 3) 10 pointsA concave mirror has a focal lengthf=21 cm.Determine the conditions on the object po-sitionpfor the image to be erect atq.Mto be positive is thatp < f.021(part 2 of 3) 10 pointsDetermine the object positionpfor the imageto be erect and 4 times the size of the object.9.f < pIn order for the image to be erect, the mag-nificationM=-qpmust be positive.(p=object distance,q= image distance).Us-ing the mirror equation, we can solve forqinterms ofpandf.1p+1q=1f,so1q=1f-1p1q=p-ff p.Now, plugging into the equation forMweobtainM=-qp=ff-p.So the condition forMto be positive is thatp < f.021(part 2 of 3) 10 pointsDetermine the object positionpfor the imageto be erect and 4 times the size of the object.What is the radius of curvature of the mirror?Assume the mirror is spherical.keywords:023(part 1 of 3) 10 pointsHint:Construct a ray diagram.
Drews, Scott – Homework 12 – Due: Nov 26 2006, midnight – Inst: Ditmire10Given:A real object is located at “p1=116f” to the left of a convergent lens with afocal lengthfas shown in the figure below.0116ff(×f)The image is1.real, erect, and smaller.2.virtual, inverted, and smaller.3.virtual, erect, and larger.4.virtual, erect, and smaller.5.virtual, inverted, and larger.6.real, inverted, and larger.correct7.real, inverted, and smaller.8.real, erect, and larger.Explanation:Basic Concepts:1p+1q=1fm=h0h=-qpConverging Lensf >0> p > ff < q <0> m >-∞f > p >0-∞< q <0> m >1Solution:

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Term
Fall
Professor
HOFFMAN
Tags
Physics, Work, Snell s Law, Correct Answer, Total internal reflection, Geometrical optics

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