# How does change in medium affect object distance/image distance?

Say,we have a container filled with a liquid of refractive index $7/5$ upto a height up $H$. There exists a plane glass mirror at the bottom of the container.

Now if a fish were placed at a height of $H/2$ from the bottom, at what distance from the fish would its image be formed?

My question is , while applying the mirror formula in this question, do we take the "object distance" to be $H/2$, that is, its actual distance from the mirror or its "optical distance" from the mirror, that is $\frac 7{10} H$?

Same for the image distance. The concept of "image" of a point means: the reflected/refracted light rays from it seem to come from the image point. So if you follow them backwards where they came from and prolongate them beyond the spot where they were reflected/refracted, they meet at the image point. Note: tt is not obviuos, that such a point must exist, the rays don't have to meet, and often they do only in some approximation.

The case of a mirror is a case, in which the image does definitely exist.
The rays are reflected following the rule, that the incident and outgoing angles are equal, always, independent of what media are involved!
The rays all seem to come from the reflected point.
This is a geometrical statement. That's why the concepts of "reflection" in the geometrical sense and the optical sense coinside. Those are genuinely different concepts, but the geometrical ("move the point in perpendicular direction by twice its distance") always implies the optical ("light ray from observer to reflected point defines a spot of reflexion; the angles of the rays spot-observer and spot-object are equal").

Now if you look from above the water, you see all distances in water smaller then they are (and behind the mirror there is water, too - the whole universe gets reflected). That is, the fish seems to be $\frac 5{14} H$ below the surface, the reflected fish $\frac{15}{14} H$ below the surface - and the spot of reflection sitll in the middle.

Note, that the distances seem smaler in water, not greater as you wrote. This is also a case of the concept of "image", but this time aproximate!
It's only valid for small objects when all light rays are nearly perpendicular.
Light rays are refracted to be neerer the perpendicular direction inside the wter; so the prolonged light rays seem to meet nearer the surface than the actual ones.

• why do you say the medium behind the mirror is also the same liquid? – Cyka Apr 29 '16 at 16:59
• because I wanted to illustrate the concept, that the whole world is reflected, and you look at it through the mirror as through a window... But what it means is: there is no surface on which rays are refracted. There is no change of medium "behind" the mirror. Try to visualize the rays yourself (also in the case of refraction on the water-air-surface), and you will see what I mean by the concept of "image", it's important to understand. – Ilja Apr 29 '16 at 17:03
• so regardless of the medium, the image and object distance remains equal? – Cyka Apr 29 '16 at 17:08
• okay, I get what you mean,thanks, and said 7/10 H instead of 5/14H only because i was talking from the sense of optical path/distance , not the apparent depth concept, but i see now why that was wrong, thank you. – Cyka Apr 29 '16 at 17:14

It depends on from which material you are viewing. If you are viewing with your eye placed inside the water then you have to consider the "object" distance because now, only the phenomena reflection takes place. However if your eye is placed above the water, then "optical" distance is taken into consideration since both reflection(from the mirror) as well as refraction(due to air-water interface) take place. Please feel free to ask for more detailed explanation.

• So if i am the fish, the image is formed at a distance H from me, right? – Cyka Apr 29 '16 at 16:58
• Exactly. Because if you are the fish then this is analogous to the condition when you see yourself in front of a mirror in real life – Zeeshan Ali Apr 29 '16 at 17:10