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Asked by Prashant DIGHE | 30 Jul, 2019, 08:15: AM

Expert Answer

If f is focal length, for real image we write the lens equation :-

begin mathsize 12px style 1 over v subscript 1 space plus space 1 over 16 space equals space 1 over f end style

...................(1)

where v₁ is the lens-to-image distance

For virtual image , if v₂ is the lens-to-image distance, we write the lens equation :-

begin mathsize 12px style 1 over v subscript 2 space plus space 1 over 6 space equals space 1 over f end style

......................(2)

if magnification is same, then we have, -(v₁/16) = v₂/6 or (1/v₂) = (-8/3)1/v₁ ...................(3)

from eqn.(1) and (2), we write,

begin mathsize 12px style 1 over v subscript 1 space plus space 1 over 16 space equals space 1 over v subscript 2 space plus space 1 over 6 space space space space space space space space space o r space space space space space space space space space space space space 1 over v subscript 1 space minus space 1 over v subscript 2 space equals space 5 over 48 end style

...........................(4)

Let us substitute (1/v₂) in eqn.(4) using eqn.(3), we get,

begin mathsize 12px style 1 over v subscript 1 plus 8 over 3 1 over v subscript 1 space equals space 5 over 48 space space space space space o r space space space space space v subscript 1 space equals space 176 over 5 space c m end style

......................(5)

By substituting for v₁ using eqn.(5), we get focal length f from eqn.(1) as ,

begin mathsize 12px style 5 over 176 space plus space 1 over 16 space equals space 1 over f space space space space space space o r space space space space space space space f space equals space 11 space c m end style

Answered by Thiyagarajan K | 31 Jul, 2019, 08:52: AM

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