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Why do we take radius as half of focal length in this problem ?

Asked by nipunverma59 | 24 Apr, 2019, 04:05: PM

Expert Answer

The focal length is given by

1 over f equals open parentheses mu minus 1 close parentheses open parentheses 1 over R subscript 1 minus 1 over R subscript 2 close parentheses

in case of plano convex lens

R subscript 1 equals infinity space a n d space R subscript 2 equals negative R a n d I f space w e space a s s u m e space mu equals 1.5 comma space a s space m o s t space o f space t h e space l e n s space i s space m a d e space u p space o f space g l a s s space h a v i n g space r e f r a c t i v e space i n d e x space a r o u n d space 1.5 T h e r e f o r e comma 1 over f equals open parentheses 1.5 minus 1 close parentheses open parentheses 1 over infinity minus fraction numerator 1 over denominator negative R end fraction close parentheses O n space s o l v i n g space w e space g e t comma f equals 2 R

But in this question we are given that it is biconvex lens.

Hence

1 over f equals open parentheses 1.5 minus 1 close parentheses open parentheses 1 over 20 minus fraction numerator 1 over denominator negative 20 end fraction close parentheses f equals 20

Now,

1 over v minus 1 over u equals 1 over f fraction numerator begin display style 1 end style over denominator begin display style v end style end fraction minus fraction numerator begin display style 1 end style over denominator begin display style negative 30 end style end fraction equals fraction numerator begin display style 1 end style over denominator begin display style 20 end style end fraction fraction numerator begin display style 1 end style over denominator begin display style v end style end fraction equals fraction numerator begin display style 1 end style over denominator begin display style 20 end style end fraction minus fraction numerator begin display style 1 end style over denominator begin display style 30 end style end fraction v equals 60

For lens,

h subscript I over h subscript o equals v over u fraction numerator begin display style h subscript I end style over denominator begin display style 2 end style end fraction equals fraction numerator begin display style 60 end style over denominator begin display style negative 30 end style end fraction h subscript i equals negative 4 c m

Hence the image formed is real and inverted having height 4 cm.

Answered by Utkarsh Lokhande | 25 Apr, 2019, 01:18: PM

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