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Solve all the three paragraph question in detail

Asked by sarveshvibrantacademy | 04 May, 2019, 10:38: AM

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)

In such a case, even with convex nature, v is negative and real image wil not be formed

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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),

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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 > [ μ₁R /( μ₂ - μ₁) ], v is negative and

virtual image is formed for the position x satisfying this condition

(B) if μ₂ < μ_{1 ,}RHS of eqn.(3) is positive, hence v is positive. Virtual image will not be formed in such a case

Answered by Thiyagarajan K | 06 May, 2019, 10:19: AM

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