# Why is the image projected onto the back of the

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Why is the image projected onto the back ofthe retina in your eyes upside down?1.The eye is a diverging lens.2.None of these3.The focal length of the eye is very long.4.The eye is a converging lens.correctExplanation:Your eye is a converging lens; your braininterprets the image as right side up.02610.0pointsA physics student places an object 6.0 cmfrom a converging lens of focal length 9.0 cm.What is the magnitude of the magnificationof the image produced?1.|M|= 3.62.|M|= 0.63.|M|= 2.04.|M|= 1.55.|M|= 3.0correctExplanation:Let :p= 0.6 cmandf= 9.0 cm.Using1p+1q=1f,q=11f-1p=119.0 cm-16.0 cm=-18.0 cm.The magnitude of the magnification is thus|M|=vextendsinglevextendsinglevextendsinglevextendsingleqpvextendsinglevextendsinglevextendsinglevextendsingle=18.0 cm6.0 cm= 3.0.
mittal (im5936) – reflection refraction imaging HW – yeazell – (58010)902710.0pointsAn ordinary magnifying glass produces animage which isThe magnification isM=-qp=--11.424 cm35.7 cm= 0.32.Theimageisvirtual,uprightand11.424 cm in front of the lens.029(part2of3)10.0pointsFind the image distance for an object distanceof 19.9 cm.
4.virtual and inverted5.real and erectExplanation:A magnifying glass is a double convex lens.The object is located just inside the focallength. Thus the image is virtual, erect, andthe magnification is larger than 1.028(part1of3)10.0pointsA diverging lens has a focal length of 16.8 cm.2f2fffLocate the image for object distance of35.7 cm .Correct answer:-11.424 cm.Explanation:Let :p= 35.7 cmandf=-16.8 cm.2ffFrom the thin lens equation1p+1q=1fq=fpp-f=-(16.8 cm) (35.7 cm)35.7 cm-(-16.8 cm)=-11.424 cm.Correct answer:-9.10954 cm.Explanation:Let :p= 19.9 cm.q=-(16.8 cm) 19.9 cm19.9 cm-(-16.8 cm)=-9.10954 cm.The magnification isM=-qp=--9.10954 cm19.9 cm= 0.457766.Theimageisvirtual,uprightand9.10954 cm in front of the lens.030(part3of3)10.0pointsFind the image distance for an object distanceof 6.9 cm.Correct answer:-4.89114 cm.Explanation:Let :p= 6.9 cm.q=-(16.8 cm) 6.9 cm6.9 cm-(-16.8 cm)=-4.89114 cm.The magnification isM=-qp=--4.89114 cm6.9 cm= 0.708861.
mittal (im5936) – reflection refraction imaging HW – yeazell – (58010)10Theimageisvirtual,uprightand4.89114 cm in front of the lens.031(part1of4)10.0pointsAn object is placed 10 m before a convexlens with focal length 6.1 m.Another concavelens is placed 11.4 m behind the first lens witha focal length-2.2 m (see the figure below).Note:Make a ray diagram sketch in orderto check your numerical answer.0510152025p1f1f110 m6.1 mf1=f2f211.4 m-2.2 mf2=At what distance is the first image from thefirst lens?Correct answer: 15.641 m.Explanation:BasicConcept:The thin lens equation is1p+1q=1

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Term
Spring
Professor
RITCHIE/LANG
Tags
refractive index, Total internal reflection
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