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The gaussian formula for a single spherical surface is defined as

$$ \frac{n}{s} + \frac{n'}{s'} = \frac{n'-n}{R}$$

When the radius of curvature, $R$, is infinite - as in the case of a planar surface, we are delt with the case

$$ \frac{n}{s} = \frac{-n'}{s'} $$

and thus

$$ s' = \frac{-sn'}{n} $$

Which means that if you look at an object in another medium with an incident angle of 0, the object may look closer or further away.

I understand why an object is not where "it seems" when looking into a different medium at an incident angle, but I dont understand why an object would appear farther or closer without and incident angle.

Why would an object appear farther or closer due to refraction when the incident angle is zero?

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The object is not 0 dimensional. Only a thin line of points lies along the line of 0 incidence angle. The rest of the object is slightly off that line, making the incident angle nonzero. This causes refraction and can make objects appear closer or farther than they are.

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    $\begingroup$ A one dimensional object, say an ideal point source of light, would also have an image in a different location than the object. $\endgroup$ – BMS Aug 30 '17 at 1:57
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Let's say you only have one eye open and you're directly in front of the object. The thing is, your pupil has some size to it. Rays that enter are not only leaving the medium surrounding the object at 0 degrees. Your lens must change shape for the rays entering the edges of the pupil to form an image on your retina, and the particular shape you force your lens into provides feedback on the distance.

Less eye-centric: An observer will see an image located at the place where diverging rays appear to diverge from. It is common to think of the image existing without the observer, but in the end some type of optical device is needed if one wants to measure the location of an image. An object embedded in a medium will emit rays in all directions. Even if you imagine an observer or other device directly in front of the object, the apparent location can be determined by back tracing two or more diverging rays, because the observer or optical device isn't sampling just a single ray that's at 0 incidence. Lenses have size. It's not just a single ray, as you seem to be assuming. Take all the rays entering the optical device, trace them back, and you'll see the image and object are at different locations.

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  • $\begingroup$ Except this is the behavior of objects regardless of observer. The image really is a different distance away than the object, and the observer just notices that. $\endgroup$ – Johnathan Gross Aug 29 '17 at 22:52
  • $\begingroup$ The image is where the diverging rays appear to have diverged from. There is nothing special here about the eye, I agree. Simply used it as an example. $\endgroup$ – BMS Aug 30 '17 at 2:04

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