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Solve all the three paragraph question in detail
question image
Asked by sarveshvibrantacademy | 04 May, 2019, 10:38: AM
answered-by-expert Expert Answer
For the refraction at spherical surface, we have     begin mathsize 12px style fraction numerator mu subscript 1 over denominator negative x end fraction plus mu subscript 2 over v space equals space fraction numerator mu subscript 2 minus mu subscript 1 over denominator R end fraction space space o r space fraction numerator begin display style mu subscript 2 end style over denominator begin display style v end style end fraction space equals space fraction numerator begin display style mu subscript 2 minus mu subscript 1 end style over denominator begin display style R end style end fraction plus space fraction numerator begin display style mu subscript 1 end style over denominator begin display style x end style end fraction end style .............................(1)
Qn. (25)
(A) as per eqn.(1),  since x and R are positive, if μ2 > μ1 , v is always positive. Hence real image is formed always if μ2 > μ1.
 
(B) false as per the explanation given above for (A)

(C) if μ2 < μ1 first term of RHS of eqn.(1) is negative. Hence if μ2 < μ1 and ( μ1 - μ2)/R > μ1/x  v will be negative. 
In such a case, even with convex nature, v is negative and real image wil not be formed
 
-----------------------------------------------
Qn.(26)
(A) as discussed in qn.(25) if μ2 < μ1 and ( μ1 - μ2)/R > μ1/x  v will be negative. if v is negative virtual mage is formed
 
if the above condition is rewritten, we have [ x > μ1 R/ ( μ1 - μ2) ] ..............(2)
Virtual image is formed only for the position of x that satisfies eqn.(2), 
 
-----------------------------
Qn.27
for a concave spherical surface, we have begin mathsize 12px style fraction numerator mu subscript 1 over denominator negative x end fraction space plus space mu subscript 2 over v space equals space fraction numerator mu subscript 2 minus mu subscript 1 over denominator negative R end fraction space space space o r space space fraction numerator begin display style mu subscript 2 end style over denominator begin display style v end style end fraction space equals space fraction numerator begin display style mu subscript 1 minus mu subscript 2 end style over denominator begin display style R end style end fraction space space plus fraction numerator begin display style mu subscript 1 end style over denominator begin display style x end style end fraction space end style ...................(3)
(A) if μ2 > μ1 , first term of RHS of eqn.(3) is negative. Hence if x > [ μ1 R /( μ2 - μ1) ], v is negative  and
virtual image is formed for the position x satisfying this condition
 
(B) if μ2 < μ1 , RHS of eqn.(3) is positive, hence v is positive. Virtual image will not be formed in such a case
 
(C) as explained in (B), if μ2 < μ1, real image is formed for any value of x
Answered by Thiyagarajan K | 06 May, 2019, 10:19: AM

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