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The adjacent figure shows a thin plano-convex lens of refractive index $μ_1$ and a thin plano-concave lens of refractive index $μ_2$, both having same radius of curvature $R$ of their curved surfaces. The thin lens of refractive index $μ_3$ has radius of curvature $R$ of both its surfaces. This lens is so placed in between the plano-convex and plano-concave lenses that the plane surfaces are parallel to each other. The focal length of the combination is

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Provided solution :

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My doubt: In the lens diagram, the medium on both the sides of the lenses are different. So I took the first one as $(\frac{µ_1}{µ_{air}}-1)$ second as $(\frac{µ_2}{µ_{1}}-1)$ and third as $(\frac{µ_3}{µ_{2}}-1)$ What is wrong with my approach? Why do we not account for the different media ?

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  • $\begingroup$ What do you mean by "first one"? $\endgroup$
    – SarGe
    Aug 11, 2020 at 8:01

1 Answer 1

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You need to understand that the formula for focal length (involving relative refractive index term)used is only valid when the medium on both sides of the lens is same though different from that of lens. In order to solve these kind of problems imagine thin slices of air in between the lenses making the formula valid. However a more true and rigorous although lengthy approach would be to consider refraction at each of the surfaces individually applying $\frac{n_2}{v} -\frac{n_1}{u}=\frac{n_2-n_1}{R}$.

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  • $\begingroup$ please use mathjax $\endgroup$
    – hyportnex
    Aug 11, 2020 at 22:47

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