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Asked by sarveshvibrantacademy | 28 Apr, 2019, 02:36: PM
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For refraction at spherical surface that separates two optical medium, we have  begin mathsize 12px style n subscript 2 over v space minus space n subscript 1 over u space equals space fraction numerator n subscript 2 minus n subscript 1 over denominator R end fraction end style ...................(1)
where n2 and n1 are refractive index of denser medium and rarer medium respectively.
R is radius of curvature of spherical surface, v is image distance and u is object distance.
Here nose of the child is object and fish is observer.
 
hence eqn.(1) is written as,   begin mathsize 12px style fraction numerator begin display style bevelled 4 over 3 end style over denominator v end fraction space minus space fraction numerator 1 over denominator negative R end fraction space equals space fraction numerator begin display style 4 over 3 minus 1 end style over denominator R space end fraction end style  .......................(2)
( Note that in eqn.(2) sign convention is applied so that the object distance u is negative) from eqn.(2) we get v = -2R
 
Hence image is seen by fish at a distance 2R from spherical surface on the same side where child is standing.
 
hence the image distance from centre of sphere = R+2R = 3R
Answered by Thiyagarajan K | 28 Apr, 2019, 04:00: PM

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