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Home / Doubts and Solutions/JEE/Physics

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Asked by sarveshvibrantacademy 4th May 2019, 10:29 AM
Answered by Expert
Answer:
For refracrion at spherical surface, with respect to the given figure, we have begin mathsize 12px style n subscript a over d subscript 1 plus n subscript g over v space equals space fraction numerator n subscript g minus n subscript v over denominator R end fraction end style  ................(1)
In eqn.(1) sign convention is not applied. v is image distance in refracting medium,  measured from point p.
R is radius of curvature which is 60 cm.
 
Let us analyze each case
 
case-(A) :-  eqn.(1) is written as  (-1/120) + (1.5/v) = (0.5/60)   or we get  v = 60 cm.
 
i.e image after refraction is formed at a distance 60 cm from P inside refracting medium.
 
Now if this image is acting as object for second refraction then we have   [ (-3/2)/60 ] + 1/v = 0.5/(-60)   or we get v = 60 cm.
 
Hence final image I1 is not formed on O which is 120 cm left of P , irrespective of distance d2

 
case-(B) :- eqn.(1) is written as (-1/240)+(1.5/v) = 0.5/60   or we get v = 120 cm
 
i.e image after refraction is formed at a distance 120 cm from P inside refracting medium.
 
Now if this image is acting as object for second refraction then we have   [ (-3/2)/120 ] + 1/v = 0.5/(-60)   or we get v = 240 cm.
 
Hence final image I1 is formed on O , but the distance d2 should be greater than 120,
otherwise mirror reflection will alter the location of first image and final image will not be formed at O
 
case-(C) and (D) are false as per the explanataion given for case -(B)
Answered by Expert 5th May 2019, 3:52 PM
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