# Optics for a flat piece of glass

An object is seen through a flat piece of glass of index of refraction 1.5 and thickness 6 cm. If the object is a distance 10 cm away form the near-surface of the glass, what is the location of the image seen by the observer directly at the other side of the glass?

I'm having trouble figuring out where the rays of light converge. By my count, any ray that goes from the arrow straight through the glass will simply re-emerge at the same angle. Since by geometry the second angle of incidence will be the same as the first angle of refraction. Therefore, if I draw two rays going straight from the top of the object to the bottom, I will simply get two parallel rays that never converge.

What rays do I draw to correctly find the location of the image? My understanding is that I should generally use a parallel line and one through the focal point, but in this case there is no focal point.

The point here (if I understand the question correctly) is to treat the eye as a point object located exactly at the edge of the glass, as in the added figure, and project the rays that hit the eyes to the adequate distance, provided that the size of the object is known to the observer.

That is to say if the observer is familiar with the object, he will perceive the object as if it is placed somewhere else.

Then it's just a matter of plugging in Snell's law which seems like you know how to apply.

• Should the magnification of the image be 1? I am still unsure what rays I should draw to determine where they intersect. Mar 30, 2017 at 5:50
• Correct. Imagine I'm showing you a familiar object, say an apple, at some distance $L$. Then I tell you to close your eyes, and when you open them the apple appears smaller. You would conclude that it is now placed further away from you, and calculate the new distance by the known size of the apple and the solid angle it takes on your field of vision. This experiment is indistinguishable to you from the one where while your eyes were closed I instead placed a crystal clear glass you are completely unaware of between you and the apple, such that the rays reach your eyes with the same angles.
– Yoni
Mar 30, 2017 at 6:34
• I'm not seeing how to determine the angle though. If I were to draw a ray from the top of the object to anywhere else the angle would be wholly arbitrary since its dependent on the height of the object, which is not given in the problem. Then my only other option is to draw a parallel ray, but if I extrapolate the angle of refraction for the parallel ray, it immediately becomes equal in height since the incident ray entered at the height of the object. Mar 30, 2017 at 6:41
• It's okay, because in the experiment I described above you knew the size of the object and by triangulating with the solid angle you experience you calculated the distance. Here instead you are given the length, so you can derive its size again by the solid angle. Now placing the piece of glass is the same as before. An extra complication is due to the fact that you don't get the luxury of witnessing the object before the glass is placed, but the physics remain the same - for any $X$ sized object, the above derivation holds.
– Yoni
Mar 30, 2017 at 6:48
• Could you clarify what angle I should be using to figure out the distance? I just don't understand where the angle even comes from. Presumably you find it via Snell's Law but in order for you to do that you need to know the initial angle which is itself dependent on the height. Mar 30, 2017 at 6:53

You are correct that the emerging rays are parallel to the incoming ones, but sketch the situation. A ray cone $POQ$ converging to point $O$ with no glass plate converges, instead, in the presence of the glass plate to point $O^\prime$ owing to the Snell's-law-induced "kink" through the glass. Indeed, the point $O^\prime$ actually depends on the angle of incidence, so that, strictly speaking, a perfect ray cone no longer converges to a point in the presence of the flat plate. Your question is presumably asking for the small angle limit calculation of $O^\prime$. This nonuniqueness of convergence point is the origin of spherical aberration, and the reason why high NA microscope objectives either have to be used with a known thickness coverslip (the compensation for the spherical aberration from that thickness is built into the design) or else they have a correction collar, allowing the user to impart a variable, compensating spherical aberration to the light field.