geometric-optics
It's written that AO is real depth and Ao1 is apparent depth. But it is looking like apparent depth is greater than real depth in the figure when oject is seen from other side of the slab. Is it possible. Can apparent depth be greater than real depth.
And I have one more doubt … In this case Real depth / Apparent depth = 1/refractive index but in general ratio of real and apparent depth equals refractive index… Why in this case this is opposite.
Thanks.
Gah. Once again, I see this question crop up because these textbooks/etc draw inaccurate diagrams.
With ONE ray, your eye can never determine where the object is. Note that in your diagram, the eye can deduce the line along which the apparent image is, but to make a point, we need two lines! Who told the eye that the apparent object was directly above the real object? So there has to be a second ray of light.
The actual diagram should be something like this:
Ok, in this diagram, we have a stick(long object) instead of a fish, and the angle $\theta$ is not small enough for the apparent image to be directly above the object, but my pount here is that in taking RD/AD , we have to take two rays. Taking one ray works only while calculating it, but it is not the reality.
Second thing: To judge depth, we need two eyes. With one eye, we can only judge depth if we know the size of the object beforehand. The second eye can be thought of as picking up the 'second ray' in the above diagram. Now there will always be some angle between alteast one eye and the normal, so refraction will still happen.
To try this out, close one eye. Now hold your arms straight out, nearly stretched, but not completely stretched. Now extend your index fingers (there should be about one inch between them). Now try touching them together. Try this a few times, then repeat with both eyes open. You'll realise how necessary the second eye is for judging depth.
While considering optical lengths and RD/AD, we always consider near-normal viewing. This means that the angle $\theta$ is very small, but not exactly 0. This allows us to draw diagrams like the one in your question to find apparent depth by considering only one ray.
Another way to look at it is to limit $\theta$ to zero in your formula for apparent depth. Surprise! No change, since there is no $\theta$ term in the final formula anyways.
So, summing up:
Oh and the answer to the question is 8 m/s, as you get $v_{app,\perp}=\frac{v_{real,\perp}}{n}$ from differentiating the apparent depth formula.
Best Answer
To answer the question in your title : Yes. Apparent depth is usually less than real depth because you are looking from a medium of lower optical density (air) into a medium of higher optical density (water or glass). If you were under water looking up at an object outside of the water, it would appear further away.
In the diagram, $O_1$ is the image of object $O$ when viewed from inside the glass block. It is located by tracing back rays inside the glass block. The object is in air, the observer is in glass, so this is a case in which apparent depth should be greater than real depth.
$I$ is the image of object $O$ when viewed from air on the other side of the glass block. It is located by tracing back rays in air in the region of the observer. Both object and image are in air, but there is a layer of optically denser material in between, so the apparent depth is again less than the real depth. However, the ratio of real to apparent depth is not as small as it would be if $O$ were inside the block.
Your formula does not work in this case because the ratio also depends on the thickness of the glass block. It is correct when there is 1 interface between different optical media - here there are 2 interfaces.
The order of the media is also important. The refractive index is more properly given not only for 2 media but also for the direction in which rays are travelling : ie air-to-glass or glass-to-air. For this reason the index is sometimes written as $$_a n_g = \frac{1}{_g n_a}$$