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A man with a near pointof 25 cm reads a book small print using a maagnifying glass, a convex lens of focal length 5 cm. (a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass? (b) what is themaximum and the minimum magnification(magnifying power) possible using the above simple microscope? |
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Answer» Solution :D = 25 cm, f = 5 cm, For closest object distance, U, the image distance, v is, - 25 cm. (near point focusing) For FARTHER object distance, (u): the corresponding distance (v.) is, `v = oo`(normal focusing) (a) To find closest IMAGEDISTANCE, lens equaiton, `(1)/(v)-(1)/(u)=(1)/(f)` Rewriting for closest object distance, `(1)/(u)=(1)/(v)-(1)/(f)` Substituting, `(1)/(u)=(1)/(-25)-(1)/(5)=(1)/(25)-(1)/(5)=((-1-5)/(25))=-(6)/(25),` `u = (25)/(6) = - 4.167 cm` The closest distance at which the person should can KEEP the book is, u = - 4.167cm. To find farthest object distance, lens equation is, `(1)/(v.)-(1)/(u.)=(1)/(f.)` Rewriting for farthestobjectdistance, `(1)/(u.) = (1)/(v.) - (1)/(f.)` Substituting, `(1)/(u.)=(1)/(oo)-(1)/(5),u=-5cm` The farthst distance at which the person can keep the book is, u = - 5 cm (b) To find magnification in near point focusing, `m=1+(D)/(f)=1+(25)/(5)=6` To findmagnification in normal focusing, `m=(D)/(f)=(25)/(5)=5` |
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