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Asked by sarveshvibrantacademy | 07 May, 2019, 12:01: PM
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Let D be the distance of the object from the pole
Then image formed is at a distance D-10 from the pole
 
As we know 
m equals fraction numerator negative v over denominator u end fraction equals h subscript I over h subscript o
Therefore
 
fraction numerator negative open parentheses D minus 10 close parentheses over denominator D end fraction equals fraction numerator negative 1 over denominator 2 end fraction 2 D minus 20 equals D D equals 20 c m D minus 10 equals 20 minus 10 equals 10 c m
 
There
object distance u=-20
Image distance v=-10
 
fraction numerator 1 over denominator negative 10 end fraction plus fraction numerator 1 over denominator negative 20 end fraction equals 1 over f f equals fraction numerator negative 20 over denominator 3 end fraction R equals 2 f T h e r e f o r e R equals negative 40 divided by 3 D i s tan c e space b e t w e e n space r a d i u s space o f space c u r v a t u r e space a n d space o b j e c t space i s space g i v e n space b y equals 20 minus 40 over 3 equals 20 over 3 c m
The distance between the pole and object is object distance which is equal to 20cm
Answered by Utkarsh Lokhande | 07 May, 2019, 12:39: PM
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