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A man with a near point of 25 cm reads ...

A man with a near point of 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 the maximum and the minimum magnification (magnifying power) possible using the above simple microscope?

Text Solution

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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 image distance, 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 farthest object distance, `(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 find magnification in normal focusing, `m=(D)/(f)=(25)/(5)=5`
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