Ray tracing is a powerful visual approach for understanding image formation, but it doesn’t provide precise information about the image. We need to develop a quantitative relationship between the object distance $s$ and the image distance $s\prime$.

To begin, FIGURE 34.38 shows a spherical boundary between two transparent media with indices of refraction $n_1$ and $n_2$. The sphere has radius of curvature $R$. Consider a ray that leaves object point P at angle $\alpha$ and later, after refracting, reaches point $\text{P}\prime$. Figure 34.38 has exaggerated the angles to make the picture clear, but we will restrict our analysis to paraxial rays traveling nearly parallel to the axis. For paraxial rays, all the angles are small and we can use the small-angle approximation.

FIGURE 34.38 Image formation due to refraction at a spherical surface. The angles are exaggerated.

The ray from P is incident on the boundary at angle ${\theta }_1$ and refracts into medium $n_2$ at angle ${\theta }_2$, both measured from the normal to the surface at the point of incidence. Snell’s law is $n_1\text{sin}{\theta }_1=n_2\text{sin}{\theta }_2$, which in the small-angle approximation is

You can see from the geometry of Figure 34.38 that angles $\alpha$, $\beta$, and $\varphi$ are related by

Using these expressions in Equation 34.15, we can write Snell’s law as

This is one important relationship between the angles.

The line of height $t$, from the axis to the point of incidence, is the vertical leg of three different right triangles having vertices at points P, C, and $\text{P}\prime$. Consequently,

But $d\to 0$ for paraxial rays, thus

This is the second important relationship that comes from Figure 34.38.

If we use Equations 34.19 in Equation 34.17, the $t$ cancels and we find

Equation 34.20 is independent of angle $\alpha$. Consequently, all paraxial rays leaving point P later converge at point $\text{P}\prime$. If an object is located at distance $s$ from a spherical refracting surface, an image will be formed at distance $s\prime$ given by Equation 34.20.

Equation 34.20 was derived for a surface that is convex toward the object point, and the image is real. However, the result is also valid for virtual images or for surfaces that are concave toward the object point as long as we adopt the sign convention shown in TABLE 34.2.

Section 34.4 considered image formation due to refraction by a plane surface. There we found (in Equation 34.13) an image distance $s\prime =-(n_2/n_1)s$. A plane can be thought of as a sphere in the limit $R\to \text{∞}$, so we should be able to reach the same conclusion from Equation 34.20. Indeed, as $R\to \text{∞}$, the term $(n_2-n_1)/R\to 0$ and Equation 34.20 becomes $s\prime =-(n_2/n_1)s$.

EXAMPLE 34.10 Image formation inside a glass rod

One end of a 4.0-cm-diameter glass rod is shaped like a hemisphere. A small lightbulb is 6.0 cm from the end of the rod. Where is the bulb’s image located?

MODEL Model the lightbulb as a point source of light and consider the paraxial rays that refract into the glass rod.

VISUALIZE FIGURE 34.39 shows the situation. $n_1=1.00$ for air and $n_2=1.50$ for glass.

FIGURE 34.39 The curved surface refracts the light to form a real image.

FIGURE 34.39 Full Alternative Text

SOLVE The radius of the surface is half the rod diameter, so $R=2.0\text{cm}$. Equation 34.20 is

$${\displaystyle \frac{1.00}{6.0\text{cm}}}+{\displaystyle \frac{1.50}{s\prime }}={\displaystyle \frac{1.50-1.00}{2.0\text{cm}}}={\displaystyle \frac{0.50}{2.0\text{cm}}}$$

Solving for the image distance $s\prime$ gives

$$\begin{array}{l}{\displaystyle \frac{1.50}{s\prime }}={\displaystyle \frac{0.50}{2.0\text{ cm}}}-{\displaystyle \frac{1.00}{6.0\text{ cm}}}=0.0833{\text{ cm}}^{-1}\\ s\prime ={\displaystyle \frac{1.50}{0.0833}}=18\text{ cm}\end{array}$$

REVIEW This is a real image located 18 cm inside the glass rod.

EXAMPLE 34.11 A goldfish in a bowl

A goldfish lives in a spherical fish bowl 50 cm in diameter. If the fish is 10 cm from the near edge of the bowl, where does the fish appear when viewed from the outside?

MODEL Model the fish as a point source and consider the paraxial rays that refract from the water into the air. The thin glass wall has little effect and will be ignored.

VISUALIZE FIGURE 34.40 shows the rays refracting away from the normal as they move from the water into the air. We expect to find a virtual image at a distance less than 10 cm.

FIGURE 34.40 The curved surface of a fish bowl produces a virtual image of the fish.

FIGURE 34.40 Full Alternative Text

SOLVE The object is in the water, so $n_1=1.33$ and $n_2=1.00$. The inner surface is concave (you can remember “concave” because it’s like looking into a cave), so $R=-25\text{cm}$. The object distance is $s=10\text{cm}$. Thus Equation 34.20 is

$${\displaystyle \frac{1.33}{10\text{cm}}}+{\displaystyle \frac{1.00}{s\prime }}={\displaystyle \frac{1.00-1.33}{-25\text{cm}}}={\displaystyle \frac{0.33}{25\text{cm}}}$$

Solving for the image distance $s\prime$ gives

$$\begin{array}{l}{\displaystyle \frac{1.00}{s\prime }}={\displaystyle \frac{0.33}{25\text{ cm}}}-{\displaystyle \frac{1.33}{10\text{ cm}}}=-0.12{\text{ cm}}^{-1}\\ s\prime ={\displaystyle \frac{1.00}{-0.12{\text{ cm}}^{-1}}}=-8.3\text{ cm}\end{array}$$

REVIEW The image is virtual, located to the left of the boundary. A person looking into the bowl will see a fish that appears to be 8.3 cm from the edge of the bowl.