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Assuming parallel rays are incident, I basically want to derive a general equation for lenses with different mediums. Can we use the effective focal length method? Also, while deriving the equation, do we have to consider sign convention for object and image distance or just writing $u$ and $v$?

Given Equation:

$$\frac{2\mu}{d} - \frac{1}{\infty} = \frac{\mu_g -1}{R} - \frac{\mu_g -\mu}{\infty}$$

But I basically want to understand what is the proof of this given equation, how do we derive it step-by-step in a general case where we can consider the both side of lens with different radius of curvatures and different refractive indices on both sides.

I have tried to edit the question appropriately so that it is useful for others facing problems in lens with different mediums

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  • $\begingroup$ People voting to close: This is a homework question but it turns on concepts about the thin lens approximation $\endgroup$
    – mmesser314
    Jun 25 '21 at 2:34
  • $\begingroup$ the RP Photonics Encyclopedia article on Lenses does not contain the derivation you are looking for, but it may help you understand lenses. Deriving the equation is off topic here because it is a homework-lie question. It would be better to ask about concepts. The link will help with those. $\endgroup$
    – mmesser314
    Jul 8 '21 at 14:30
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You are mistaken about purpose of the thin lens approximation.

Assume a ray strikes a lens at a height $h$ from the axis. It travels through the lens at some angle $\theta$. After a horizontal distance $d$, it comes to the next surface. You can calculate the new height as $h_2 = h + d \space sin(\theta)$. The thin lens approximation means you approximate $d = 0$ and $h_2 = h$.

You would think this approximation implies the curvature of the lens is also $0$, which makes it a window. But this is not the purpose of the approximation. It is just to simplify optical calculations a bit. This simplification is only useful for a first approximation, but it helped a lot in the days before computers. And it is still useful when you are just setting up focal lengths and such of a system, before considering aberrations and the real design. Or when you have a simple single lens system and don't care about exact placement.

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  • $\begingroup$ I have understood the use of thin lens approximation from your answer, can you please explain the given solution, because I am not understanding their concept? $\endgroup$
    – PCMSE
    Jun 24 '21 at 15:21
  • $\begingroup$ You assumed the rays are undeviated when they pass through the lens because the lens is thin. This isn't right. It is a lens, and you should treat it like one. You can just pretend that the thickness is $0$, which may make your calculations easier. That is, assume your lens is a position x. When a ray gets to x, it is deviated by the front surface to a new angle. Then it immediately hits the back surface and is deviated again to another angle. $\endgroup$
    – mmesser314
    Jun 25 '21 at 2:29

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