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The question is as follow :{if an object is placed at a distance R from the pole of first surface ,then the real image is formed at a distance R from the pole of the second surface.the refractive index μ of the sphere is given by ? }                     you can use the paragraph given below
question image
Asked by sarveshvibrantacademy | 29 Apr, 2019, 10:40: AM
answered-by-expert Expert Answer
we have refraction formula for spherical surface,  begin mathsize 12px style n subscript 2 over v minus n subscript 1 over u equals fraction numerator n subscript 2 minus n subscript 1 over denominator R end fraction end style .................(1)
where n2 is the refrative index of denser medium, n1 is refrative index of rarer medium,
v is the image distance from pole of spherical surface, u is the object distance and R is radius of curvature.
 
In the question it is given that real image is formed at a distance R from the pole of second surface.
The ambiguity is whether the real image is formed inside sphere, i.e., left side of second surface or
outside sphere i.e. right side of second surface.
 
If we consider image is formed at left side of second surface, solution for μ is not arrived from eqn.(1),
hence image formation at left side is not possible.
 
If we consider image is formed at right side of second surface, then medium in sphere is acting like a medium of lens.
 
Hence we apply lens formula to get focal length f,    begin mathsize 12px style 1 over f space equals space open parentheses mu minus 1 close parentheses open parentheses 1 over R plus 1 over R close parentheses space equals space fraction numerator 2 open parentheses mu minus 1 close parentheses over denominator R end fraction end style ................(2)
since we have equation for lens,  (1/f) = (1/v) - (1/u)   ........................(3)
 
using eqn.(2) and (3) and using proper sign convention,

we write for image formation by sphere as,  begin mathsize 12px style fraction numerator 2 open parentheses mu minus 1 close parentheses over denominator R end fraction space equals space 1 over R space plus space 1 over R end style ............(4)
we get μ = 2 from eqn.(4)
Answered by Thiyagarajan K | 29 Apr, 2019, 16:05: PM

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