How tall?

Ecco's escape

To answer this question, we need to break it down into two pieces:

1. Does the hole in the wall look higher or lower from Ecco's perspective under the water?
2. Does the hole in the wall look larger or smaller from Ecco's perspective under the water?

To tackle both problems, we'll use the diagram below. The rays were drawn following Snell's Law , which governs the angle of incidence and refraction of a light ray passing from one medium with index of refraction $n_1$ to another with index $n_2$ . Note that the angles $\theta$ are with respect to the normal to the interface.

$\begin{aligned} n_1 \sin{\theta_i} &= n_2 \sin{\theta_r} \\ \frac{n_1}{n_2} \sin{\theta_i} &= \sin{\theta_r} \end{aligned}$

Check out my reply below for a derivation of Snell's Law.

Since the light rays from the hole to freedom are in air, and Ecco is in water, $n_1 = n_{\textrm{air}} = 1.00$ and $n_2 = n_{\textrm{water}} = 1.33$ . This means that $\sin{\theta_r}$ is less than $\sin{\theta_i}$ . Since the sine function only increases from $0$ to $90^{\circ}$ , this implies $\theta_r$ must be less than $\theta_i$ : The angle of refraction is less than the angle of incidence, and so the ray will bend towards the normal going from air to water.

Should Ecco aim high or low?

The apparent image of the hole as it appears to Ecco is found by extending the refracted beam back to the wall. Since the angle of refraction is less than the angle of incidence, this image appears to be coming from higher in the sky than the actual hole. Ecco will have to aim low for his jump.

Does it look like Ecco can fit?

To figure out how large the hole looks from Ecco's perspective, we'll need to look at two points on the hole. One at the top $A$ and one at the bottom $B$ . The corresponding points on the apparent image of the hole are $A'$ and $B'$ . We'll use the angle subtended by the hole in Ecco's vision as a representation of its apparent "size". So the actual size of the hole is $\Delta_i = \angle YZB - \angle WXA$ (assuming that Ecco is close to the water's surface) while the apparent size is $\Delta_r = \angle YZB' - \angle WXA'$ . This angle is just the difference of the angle coming from the bottom of the hole, to the one coming from the top of the hole. For the real size, it's the incident angles, and for the apparent size it's the refracted angles:

$\Delta_i = \theta_i^{\textrm{bottom}} - \theta_i^{\textrm{top}}\\ \Delta_r = \theta_r^{\textrm{bottom}} - \theta_r^{\textrm{top}}$

To keep things simple, let's just say that the real hole subtends an angle of $30^{\circ}$ , with the incident angle from the top measuring $30^{\circ}$ from the normal, and the incident angle from the bottom measuring $60^{\circ}$ from the normal. We can calculate their angles of refraction using the equation we used above:

$\sin{\theta_r} = \frac{n_1}{n_2} \sin{\theta_i}$

Using the inverse sine, this yields $\theta_r^{\textrm{top}} = 22 ^ \circ$ and $\theta_r^{\textrm{bottom}} = 40.5 ^ \circ$ . Since the apparent subtended angle is just their difference, this means from the point of view of Ecco, the hole subtends only $18.5 ^ \circ$ . Thus, it looks smaller, and harder to fit through than it really is.

In general

Above we just picked two random incident angles for the top and bottom of the hole. In the case we looked at, the apparent subtended angle was less than the real one. Is this always true? Let's keep things in general and find out.

The subtended angle of the actual hole is just:

$\Delta_i = \theta_i^{\textrm{bottom}} - \theta_i^{\textrm{top}}$

And the apparent subtended angle of the hole's image is:

$\Delta_r = \theta_r^{\textrm{bottom}} - \theta_r^{\textrm{top}}$

Using the first equation we derived from Snell's law, and the inverse sine, we can substitute in that expression for the angles of refraction:

$\delta_r =\arcsin \left( \frac{n_1}{n_2} \sin \theta_i^{\textrm{bottom}} \right) - \arcsin \left( \frac{n_1}{n_2} \sin \theta_i^{\textrm{top}} \right)$

There might be some way to simplify this and express it in terms of $\theta_i^{\textrm{bottom}} - \theta_i^{\textrm{top}}$ , but nobody's got time for that. Let's just plot it for all $\theta_i^{\textrm{bottom}} > \theta_i^{\textrm{top}}$ (orange) and compare it to the real hole's subtended angle (black):

The two dependent variables here are he incident angles from the top and bottom of the hole, expressed in radians.

The image's subtended angle (and size) is less than the real hole's for all incident angles (so for all positions of the hole).

As water has larger refractive index when compared to air so the light rays from the dolphin travel through the water and when they reached the water-air contact surface tue light rays bend away from the normal (since water has more refractive index and light bends away from the normal when it enters into rarer medium from denser medium)

Therefore, from the frame of dolphin the hoop looks smaller in size and higher than it really is.

Note : The options say harder to fit in which is nothing but smaller in size.

Fermat's Principle states that light takes the path between two points that minimizes time. When considering the refraction index of air and water, one must note that light moves faster through air than it does through water. So to minimize time, light travels a longer distance through the air than through the water. Now consider the line between the tip of the dolphin's snout, and the point at which light hits the water based on this least-time principle. This is, in a sense, the dolphin's perception of the hoop, where the slope of the line determines the "height" of the loop from the dolphin's perspective. Now consider the line drawn directly through the tip of the dolphin's snout, and the hoop (imagine the line passing through the center of the hoop). This is, in a sense, the accurate representation of the "height" of the hoop (this is how the dolphin would see the hoop if the water didn't exist). The slope of this line is less than the slope of the line that light actually takes, so the hoop appears higher from the dolphin's perspective.

To determine whether, from the dolphin's perspective, the hoop seems easier to fit through, we will apply Fermat's Principle once again. This time, we consider two paths, each of which represents the path that light would take to hit the dolphin from the hoop. One path runs through the top of the hoop, and the other path runs through the bottom of the hoop. Now we will draw two lines. This time, we connect the top of the loop directly to the tip of the snout, and we do the same with the bottom of the loop. We consider a hoop to appear "harder" to go through if the length of the line segment between the two points at which light hits the water from our paths.. is less than the length of the line segment between the the two points at which those two lines hit the water. Although I am not able to prove this myself, it is true that the hoop appears "harder" with these metrics. To assure myself of this fact, I applied the following heuristic:

Let f(x) be the function whose output is the length of the line segment between the two points at which light hits the water given the paths between the top/bottom of the hoop to the dolphin. The input of this function is x = n.water / n.air, the ratio of the refraction indices of water and air. If we let this ratio grow infinitely large, we see that the path traced by any point on the hoop to the dolphin goes directly from the hoop, to the point on the water directly above the dolphin, then straight down (this is to minimize the distance spent in the water). Because this applies to any point on the hoop, the points at which the paths from the bottom and the top of the hoop to the dolphin hit the water will become infinitesimally close, as those points approach the same value. So for the function f(x), as x approaches infinity, f(x) approaches 0. Now we must make an assumption: The derivative of f(x) is negative for all x < 1. This can be reasoned by letting the ratio of the refraction indices of water and air increase, and imagining what happens to the paths from the top/bottom of the hoop to the dolphin. Note that as this ratio, x, approaches infinity, f(x) approaches 0, which means that if f(x) is well defined (which we assume it is), and if there exists any c such that f(c) > 0, there must be some interval on f(x) for which the derivative of f(x) is negative. I assume the entire interval [1, inf] has the derivative of f(x) be negative because in order for the derivative of f(x) to become positive, there must be some ratio at which the aforementioned line segment starts increasing. In my analysis when solving this problem, I saw no such ratio existing.

Now with all this in mind, the calculus becomes simple. Note that when we drew the lines from the top/bottom of the hoop to the dolphin, we had the scenario where the refraction indices of the water and air are equal: where x = 1. Also note that the length of the line segment connecting the two points at which the lines hit the water is indeed some positive number (this can be evaluated geometrically using the picture). Now we have a c where f(c) > 0, because c = 1 and f(1) > 0. In the problem, we are given a ratio n.water/n.air = 1.33/1 = 1.33. Now we consider the assumption that the derivative of f(x) is negative for all x > 1, and deduce that f(1.33) < f(1). Therefore, The hoop appears harder to fit through from the dolphin's perspective.

sina/sina'=sinb/sinb'=n(water) sina'/sinb'=sina/sinb Δθ(out of water)=b-a Δθ(in water)=b'-a' then look at the unit circle:attention that sina'/sinb'=sina/sinb and that is (here I use "L" for "the length of xx color line in the left") Lpink/Lred=Lpurple/Lgreen so it's apparent that the angle between the green line and the dark blue line in the right (the verticle angle when seeing the lights above water) is bigger than the angle between the light green line and the light blue line in the right (the verticle angle when seeing the lights under water), which means an overwater object's vertical size appears to be smaller when being seen under water (sorry , I'm having trouble using my mouse, which was making me mad while I was drawing.....)

Quick answer: This is the opposite case of a pencil in a glass full of water where you see the pencil look closer and larger.

Hence, you would expect the object to look smaller and further away.