I understood the concept of apparent depth from here:

enter image description here enter image description here

But one thing I didn't understand is, will there be difference in the real depth and apparent depth when we are looking not at an angle as shown above but vertically downward (along the normal) as shown in the figure below :

enter image description here

According to me real depth and apparent depth should be same because the light rays coming out of the object is not undergoing any refraction.

If I'm right. Then the answer to the question below must be 6 m/s but Its 8m/s. How come?

A bird is flying 3 m above the surface of water. If the bird is diving vertically down with speed = 6 m/s, his apparent velocity as seen by a stationary fish underwater is : (A) 8 m/s (B) 6 m/s (C) 12 m/s (D) 4 m/s

  • 1
    $\begingroup$ What would happen if the angles were not small? Would the image be shifted horizontally too? Or will the formula will not hold true in this situation? $\endgroup$
    – Aniansh
    Feb 26, 2016 at 16:49

3 Answers 3


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:

stick in water

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:

  1. One ray is not enough to judge depth
  2. Neither is one eye
  3. So perfectly perspendicular viewing makes no sense for a two-eyed depth-judging being.
  4. So we can easily apply the formula for real depth apparent depth when the observer is on the normal.

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.

  • $\begingroup$ It must be kept in mind that even though the final formula doesn't have a $\theta$ term in it, the derivation for $\text{Real Depth/Apparent Depth}=\mu$ involves the assumption that $\theta$ is small. So the result is not valid for cases when viewed from a position quite away from the normal. $\endgroup$
    – Vishnu
    Dec 28, 2019 at 13:26
  • $\begingroup$ physics.stackexchange.com/questions/659624/… $\endgroup$ May 30, 2022 at 21:34

I did the experiment by putting a ruler into a glass of water, and I found that the perception of reduced depth remained apparent as close as vertical as I could get (too close to vertical I couldn't see the marks on the ruler).

Note that your final result for $n_w$

$$n_w = \frac{D_r}{D_a}$$

does not contain any dependence on the angle of view, so $D_a$ remains less than $D_r$ arbitrarily near the vertical. At the vertical your calculation involves zero divided by zero so we can't do the calculation there, but at an infinitesimally small angle away from vertical you get $D_a$ < $D_r$ so I would assume the result holds at zero incident angle as well. That presumably explains you fish's eye view.

Response to claw's comment:

This (rather rough!) diagram shows the bird as seen by the fish:

Fish's eye view

I'm going to use your assumption that i and r are small so i = sine(i) = tan(i) and the same for r. I'm also assuming that the distance from the water to the bird, $h$, is much greater than the distance of the fish under the water, so:

$$\frac {w}{h} = tan(i) = i$$

where $w$ is the distance from the birds body to the tip of the wing i.e. half the wingspan (I chose it to be half the wingspan to avoid messing around with factors of two). Now the fish sees the bird in the position I've drawn in red i.e. at some height, $h'$ and with some wing length, $w'$, and:

$$\frac {w'}{h'} = tan(r) = r$$

Now we know from Snell's law that sine(r) = sine(i)/n, and using our approximation that $i$ and $r$ are small we get:

$$\frac {w'}{h'} = \frac{1}{n_w} \frac {w}{h}$$

Now there's a key point to be made. Assuming the fish doesn't have binocular vision, our fish only knows w'/h' i.e. the fish can't tell if it's huge bird far away or a small bird very near. To make any progress we need to assume that the fish knows what the wingspan of the bird is, i.e. the fish knows that w' = w. If we know w' = w we can divide both sides by w and get:

$$\frac {1}{h'} = \frac{1}{n_w} \frac {1}{h}$$

or with a simple rearrangement:

$$h' = n_wh$$

And this is the result we need. The velocity seen by the fish, $v'$, is just dh'/dt:

$$v' = \frac{dh'}{dt} = \frac{d(n_wh)}{dt} = n\frac{dh}{dt} = n_wv$$

  • $\begingroup$ But there is no refraction? I mean go through the same derivation except the eye is on the vertical line. $\endgroup$
    – claws
    Feb 11, 2012 at 10:41
  • $\begingroup$ @JohnRennie Being near the vertical is not enough, that is already taken care of in the question. So, for true experimentally accurate data, you should close one eye and view it from exactly the vertical. See my answer. $\endgroup$ Feb 11, 2012 at 10:41
  • 1
    $\begingroup$ @claws: I've expanded my answer to address your comment. In brief, the fish judges the bird's distance, and therefore speed, by it's apparent size, and the apparent size is affected by refraction. NB you're correct that if everything were completely vertical the fish couldn't judge distance or speed. However that would require the bird to be infinitely small or infinitely far away, neither of which is physically realistic! $\endgroup$ Feb 11, 2012 at 14:17

$aNw = 1.33$ (refractive index of air/water interface from tables)

Conversely $wNa = \frac{1}{1.33} = 0.75$ ( going from water to air)

$Dr/Da = N$ (real depth / apparent depth = refractive index)

So, $Dr/Da = N$ becomes

$3m / Da = 0.75$ $Da = 3 / 0.75 = 4m$

The bird appears to be flying at $4m$ above the water surface.

This makes sense because we know that objects in water appear closer to the surface than they are, and therefore objects above the water surface when viewed from below would appear further away.

The bird dives at $6\, \mathrm{m/s}$ and covers the $3m$ distance in $0.5s$.

The fish sees him covering 4m but the time remains the same so he covers an apparent $4m$ in $0.5$ seconds. This gives him an apparent speed of $8\, \mathrm{m/s}$ which is the answer.


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