34.6.2 Thin-Lens Image Formation

Although the thin-lens equation allows precise calculations, the lessons of ray tracing should not be forgotten. The most powerful tool of optical analysis is a combination of ray tracing, to gain an intuitive understanding of the ray trajectories, and the thin-lens equation.

EXAMPLE 34.13 Designing a lens

The objective lens of a microscope uses a plano-convex glass lens with the flat side facing the specimen. A real image is formed 160 mm behind the lens when the lens is 8.0 mm from the specimen. What is the radius of the lens’s curved surface?

MODEL Treat the lens as a thin lens with the specimen as the object. The lens’s focal length is given by the lens maker’s equation.

VISUALIZE FIGURE 34.43 clarifies the shape of the lens and defines $R_{2}$ . The index of refraction was taken from Table 34.1.

The figure illustrates a plano-convex microscope lens.

FIGURE 34.43 Full Alternative Text

SOLVE We can use the lens maker’s equation to solve for $R_{2}$ if we know the lens’s focal length. Because we know both the object and image distances, we can use the thin-lens equation to find

\frac{1}{f} = \frac{1}{s} + \frac{1}{s'} = \frac{1}{8.0 mm} + \frac{1}{160 mm} = 0.131 {mm}^{- 1}

The focal length is $f = 1 / (0.131 {mm}^{- 1}) = 7.6 mm,$ but $1 / f$ is all we need for the lens maker’s equation. The front surface of the lens is planar, which we can consider a portion of a sphere with $R_{1} \to ∞$ . Consequently $1 / R_{1} = 0$ . With this, we can solve the lens maker’s equation for $R_{2} :$

\begin{matrix} \frac{1}{R_{2}} & = & \frac{1}{R_{1}} - \frac{1}{n - 1} \frac{1}{f} = 0 - (\frac{1}{1.50 - 1}) (0.131 {mm}^{- 1}) \\ = & - 0.262 {mm}^{- 1} \\ R_{2} & = & - 3.8 mm \end{matrix}

The minus sign appears because the curved surface is concave toward the object. Physically, the radius of the curved surface is 3.8 mm.

REVIEW The actual thickness of the lens has to be less than $R_{2}$ , probably no more than about 1.0 mm. This thickness is significantly less than the object and image distances, so the thin-lens approximation is justified.

EXAMPLE 34.14 A magnifying lens

A biologist uses a magnifying lens that sits 2.0 cm above a specimen. The magnification is 4.0. What is the focal length of the lens?

MODEL A magnifying lens is a converging lens with the object distance less than the focal length $(s < f)$ . Assume it is a thin lens.

VISUALIZE FIGURE 34.44 shows the lens and a ray-tracing diagram. We do not need to know the actual shape of the lens, so the figure shows a generic converging lens.

The figure illustrates the ray tracing diagram of a magnifying lens.

FIGURE 34.44 Full Alternative Text

SOLVE A virtual image is upright, so $m = + 4.0$ . The magnification is $m = - s' / s$ , thus

s' = - 4.0 s = - (4.0) (2.0 cm) = - 8.0 cm

We can use $s$ and $s'$ in the thin-lens equation to find the focal length:

\begin{matrix} \frac{1}{f} & = & \frac{1}{s} + \frac{1}{s'} = \frac{1}{2.0 cm} + \frac{1}{- 8.0 cm} = 0.375 {cm}^{- 1} \\ f & = & 2.7 cm \end{matrix}

REVIEW $f > 2 cm,$ as expected.

34.6.2 Thin-Lens Image Formation

EXAMPLE 34.13 Designing a lens

FIGURE 34.43 A plano-convex microscope lens.

EXAMPLE 34.14 A magnifying lens

FIGURE 34.44 Pictorial representation of a magnifying lens.