Hiding behind a goldfish

The incoming rays form an angle of (90-α) = 60 degrees with the normal, which is the angle of incidence.

Using Snell's law, we can find the angle of refraction (r): 1.33 = sin 60 / sin r hence r = 40.63 degrees.

Now, the angle of the refracted ray with the horizontal is [(90-r) - (90-i)]= (i-r) = 19.37 degrees.

Let the length of the refracted ray be L. Then L sin(i-r) is also the vertical shift = 0.5 m L cos(r) = X

Solving yield X = 1.14 m

Initially when the water enters the aquarium, it changes its direction: $\sin(90^\circ - \alpha) = n \sin(\beta)$ , where $\beta$ is the angle the light makes in the aquarium with the normal.

When the light goes out of the aquarium, the exiting angle is the same as the entering angle into the aquarium since the relative refractive index of the boundaries is the same. The length $L$ traveled by the light ray in the aquarium is $L=X/cos \beta$ . The vertical displacement of the light ray (equal to $d/2$ ) is $d/2=L sin \gamma$ where $\gamma=90-\alpha-\beta$ . Using $d=1$ yields \X=1.144).

After a geometrical consideration of the system, we can derive that the light ray shifts for $d = l \cos(\alpha) (1 - \frac{\sin(\alpha)}{n})$ , where $l$ is the width of aquarium. Since we need to hide $2d = 1~\mbox{m}$ , we can calculate $l$ to be $l = 2.8665~\mbox{m}$ .