34.7.3 The Mirror Equation

The thin-lens equation assumes lenses have negligible thickness (so a single refraction occurs in the lens plane) and the rays are nearly parallel to the optical axis (paraxial rays). If we make the same assumptions about spherical mirrors—the mirror has negligible thickness and so paraxial rays reflect at the mirror plane—then the object and image distances are related exactly as they were for thin lenses:

\begin{matrix} \frac{1}{s} + \frac{1}{s'} = \frac{1}{f} & (mirror equation) \end{matrix}

(34.27)

The focal length of the mirror, as you can show as a homework problem, is related to the mirror’s radius of curvature by

f = \frac{R}{2}

(34.28)

TABLE 34.4 shows the sign convention used with spherical mirrors. It differs from the convention for lenses, so you’ll want to carefully compare this table to Table 34.3. A concave mirror (analogous to a converging lens) has a positive focal length while a convex mirror (analogous to a diverging lens) has a negative focal length. The lateral magnification of a spherical mirror is computed exactly as for a lens:

m = - \frac{s'}{s}

(34.29)

TABLE 34.4 Sign convention for spherical mirrors

	Positive	Negative
$R$ , $f$	Concave toward the object	Convex toward the object
$s'$	Real image, same side as object	Virtual image, opposite side from object

EXAMPLE 34.16 Analyzing a concave mirror

A 3.0-cm-high object is located 20 cm from a concave mirror. The mirror’s radius of curvature is 80 cm. Determine the position, orientation, and height of the image.

MODEL Treat the mirror as a thin mirror.

VISUALIZE The mirror’s focal length is $f = R / 2 = + 40 cm,$ where we used the sign convention from Table 34.4. With the focal length known, the three special rays in FIGURE 34.51 show that the image is a magnified, virtual image behind the mirror.

The figure illustrates the image formation for a concave mirror.

FIGURE 34.51 Full Alternative Text

SOLVE The thin-mirror equation is

\frac{1}{20 cm} + \frac{1}{s'} = \frac{1}{40 cm}

This is easily solved to give $s' = - 40 cm,$ in agreement with the ray tracing. The negative sign tells us this is a virtual image behind the mirror. The magnification is

m = - \frac{- 40 cm}{20 cm} = + 2.0

Consequently, the image is 6.0 cm tall and upright.

REVIEW This is a virtual image because light rays diverge from the image point. You could see this enlarged image by standing behind the object and looking into the mirror. In fact, this is how magnifying cosmetic mirrors work.

CHAPTER 34 CHALLENGE EXAMPLE Optical fiber imaging

An endoscope is a thin bundle of optical fibers that can be inserted through a bodily opening or small incision to view the interior of the body. As FIGURE 34.52 shows, an objective lens forms a real image on the entrance face of the fiber bundle. Individual fibers, using total internal reflection, transport the light to the exit face, where it emerges. The doctor (or a TV camera) observes the object by viewing the exit face through an eyepiece lens.

A photograph of a doctor performing endoscopy on a patient.

The figure illustrates the working principle of an endoscope.

FIGURE 34.52 Full Alternative Text

Consider an endoscope having a 3.0-mm-diameter objective lens with a focal length of 1.1 mm. These are typical values. The indices of refraction of the core and the cladding of the optical fibers are 1.62 and 1.50, respectively. To give maximum brightness, the objective lens is positioned so that, for an on-axis object, rays passing through the outer edge of the lens have the maximum angle of incidence for undergoing TIR in the fiber. How far should the objective lens be placed from the object the doctor wishes to view?

MODEL Represent the object as an on-axis point source and use the ray model of light.

VISUALIZE FIGURE 34.53 shows the real image being focused on the entrance face of the endoscope. Inside the fiber, rays that strike the cladding at an angle of incidence greater than the critical angle $θ_{c}$ undergo TIR and stay in the fiber; rays are lost if their angle of incidence is less than $θ_{c}$ . For maximum brightness, the lens is positioned so that a ray passing through the outer edge refracts into the fiber at the maximum angle of incidence $θ_{max}$ for which TIR is possible. A smaller-diameter lens would sacrifice light-gathering power, whereas the outer rays from a larger-diameter lens would impinge on the core-cladding boundary at less than $θ_{c}$ and would not undergo TIR.

The figure illustrates the magnified view of the entrance of an optical fiber.

FIGURE 34.53 Full Alternative Text

SOLVE We know the focal length of the lens. We can use the geometry of the ray at the critical angle to find the image distance $s'$ , then use the thin-lens equation to find the object distance $s$ . The critical angle for TIR inside the fiber is

θ_{c} = {sin}^{- 1} (\frac{n_{cladding}}{n_{core}}) = {sin}^{- 1} (\frac{1.50}{1.62}) = 67.8 °

A ray incident on the core-cladding boundary at exactly the critical angle must have entered the fiber, at the entrance face, at angle $θ_{2} = 90 ° - θ_{c} = 22.2 °$ . For optimum lens placement, this ray passed through the outer edge of the lens and was incident on the entrance face at angle $θ_{max}$ . Snell’s law at the entrance face is

n_{air} sin θ_{max} = 1.00 sin θ_{max} = n_{core} sin θ_{2}

and thus

θ_{max} = {sin}^{- 1} (1.62 sin 22.2 °) = 37.7 °

We know the lens radius, $r = 1.5 mm,$ so the distance of the lens from the fiber—the image distance $s'$ —is

s' = \frac{r}{tan θ_{max}} = \frac{1.5 mm}{tan (37.7 °)} = 1.9 mm

Now we can use the thin-lens equation to locate the object:

\begin{matrix} \frac{1}{s} & = & \frac{1}{f} - \frac{1}{s^{'}} = \frac{1}{1.1 mm} - \frac{1}{1.9 mm} \\ s & = & 2.6 mm \end{matrix}

The doctor, viewing the exit face of the fiber bundle, will see a focused image when the objective lens is 2.6 mm from the object she wishes to view.

REVIEW The object and image distances are both greater than the focal length, which is correct for forming a real image.

34.7.3 The Mirror Equation

TABLE 34.4 Sign convention for spherical mirrors

EXAMPLE 34.16 Analyzing a concave mirror

FIGURE 34.51 Pictorial representation of Example 34.16.

CHAPTER 34 CHALLENGE EXAMPLE Optical fiber imaging

FIGURE 34.52 An endoscope.

FIGURE 34.53 Magnified view of the entrance of an optical fiber.