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Asked by sarveshvibrantacademy | 14 May, 2019, 10:00: AM
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Let us use lens makers formula to get focal length.
 
if F is focal length of given diverging lens made of material of refractive index μ,
 
then we have  (1/F) = (μ-1) [ (1 / -R) - (1 / R) ] = -2(μ-1) / R .....................(1)
 
where R is radii of curvature of the spherical surface.
 
After modifying the given lens as a combination of three lenses as given in the question, focal length F is given by
 
(1/F) = (1/f1) + (1/f2) + (1/f3)   .................  (2)
 
where f1 is the focal length of left side plano-concave lens made of material of refractive index μ,
f2 is focal length of converging lens made of liquid of refractive ndex 3 and f3 is the focal lenth of right side
plano-concave lens made of material of refractive index μ .
 
By applying lens makers formula for each lens given eqn.(2), using eqn.(1) for LHS of eqn.(2)( F is +ve)
and by considering the proper relaive refractive index for each lens,
 
we get,  begin mathsize 12px style fraction numerator 2 open parentheses mu minus 1 close parentheses over denominator R end fraction space equals space fraction numerator open parentheses mu minus 1 close parentheses over denominator R end fraction space plus space fraction numerator begin display style 2 end style open parentheses begin display style 3 over mu minus 1 end style close parentheses over denominator R end fraction plus open parentheses mu over 3 minus 1 close parentheses open parentheses negative 1 over R close parentheses end style ...........................(3)
In left side of above eqn.(3),  positive value of eqn.(1) is considered , because the lens combination is converging lens.
 
after simplification of eqn.(2),  we get    (4/3)μ - (6/μ) = 0   or μ2 = 18/4   or  μ = 3√2/2  
Answered by Thiyagarajan K | 14 May, 2019, 19:58: PM

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