If you see a fish that appears to be swimming close to the front window of the aquarium, but then look through the side of the aquarium, you’ll find that the fish is actually farther from the window than you thought. Why is this?

To begin, recall that vision works by focusing a diverging bundle of rays onto the retina. The point from which the rays diverge is where you perceive the object to be. FIGURE 34.25a shows how you would see a fish out of water at distance $d$.

FIGURE 34.25 Refraction of the light rays causes a fish in the aquarium to be seen at distance $d\prime$.

FIGURE 34.25 Full Alternative Text

Now place the fish back into the aquarium at the same distance $d$. For simplicity, we’ll ignore the glass wall of the aquarium and consider the water-air boundary. (The thin glass of a typical window has only a very small effect on the refraction of the rays and doesn’t change the conclusions.) Light rays again leave the fish, but this time they refract at the water-air boundary. Because they’re going from a higher to a lower index of refraction, the rays refract away from the normal. FIGURE 34.25b shows the consequences.

A bundle of diverging rays still enters your eye, but now these rays are diverging from a closer point, at distance $d\prime$. As far as your eye and brain are concerned, it’s exactly as if the rays really originate at distance $d\prime$, and this is the location at which you see the fish. The object appears closer than it really is because of the refraction of light at the boundary.

We found that the rays reflected from a mirror diverge from a point that is not the object point. We called that point a virtual image. Similarly, if rays from an object point P refract at a boundary between two media such that the rays then diverge from a point $\text{P}\prime$ and appear to come from $\text{P}\prime$, we call $\text{P}\prime$ a virtual image of point P. The virtual image of the fish is what you see.

Let’s examine this image formation a bit more carefully. FIGURE 34.26 shows a boundary between two transparent media having indices of refraction $n_1$ and $n_2$. Point P, a source of light rays, is the object. Point $\text{P}\prime$, from which the rays appear to diverge, is the virtual image of P. Distance $s$ is called the object distance. Our goal is to determine distance $s\prime$, the image distance. Both are measured from the boundary.

FIGURE 34.26 Finding the virtual image $\text{P}\prime$ of an object at P. We’ve assumed $n_1>n_2$.

A line perpendicular to the boundary is called the optical axis. Consider a ray leaving the object at angle ${\theta }_1$ with respect to the optical axis. ${\theta }_1$ is also the angle of incidence at the boundary, where the ray refracts into the second medium at angle ${\theta }_2$. By tracing the refracted ray backward, you can see that ${\theta }_2$ is also the angle between the refracted ray and the optical axis at point $\text{P}\prime$.

The distance $l$ is common to both the incident and the refracted rays, and you can see that $l=s\text{tan}{\theta }_1=s\prime \text{tan}{\theta }_2$. However, it is customary in optics for virtual image distances to be negative. (The reason will be clear when we get to image formation by lenses.) Hence we will insert a minus sign, finding that

Snell’s law relates the sines of angles ${\theta }_1$ and ${\theta }_2;$ that is,

In practice, the angle between any of these rays and the optical axis is very small because the size of the pupil of your eye is very much less than the distance between the object and your eye. (The angles in the figure have been greatly exaggerated.) Rays that are nearly parallel to the axis are called paraxial rays. The small-angle approximation $\text{sin}\theta \approx \text{tan}\theta \approx \theta$, where $\theta$ is in radians, can be applied to paraxial rays. Consequently,

Using this result in Equation 34.10, we find that the image distance is

The minus sign tells us that we have a virtual image.