Chapter 24, Problem 30P

Set up: $\frac{1}{s}+\frac{1}{s\prime }=\frac{1}{f}$. The type of lens determines the sign of f. $m=\frac{y\prime }{y}=-\frac{s\prime }{s}$. The sign of s′ determines whether the image is real or virtual, s = +8.00 cm. s′ = −3.00 cm. s′ is negative because the image is on the same side of the lens as the object.

Solve: (a) $\frac{1}{f}=\frac{s+s\prime }{ss\prime }$ and $f=\frac{ss\prime }{s+s\prime }$= $\frac{(8.00\text{ cm})(-3.00\text{ cm})}{8.00\text{ cm-3}\text{.00 cm}}$ = −4.80 cm. f is negative so the lens is diverging.

(b) $m=-\frac{s\prime }{s}=-\frac{-3.00\text{ cm}}{8.00\text{ cm}}$ = +0.375. y′ = my = (0.375)(6.50 mm) = 2.44 mm. s′ <0 and the image is virtual.

Reflect: A converging lens can also form a virtual image, if the object distance is less than the focal length. But in that case |s′| > s and the image would be farther from the lens than the object is.

30. Figure 24.45 shows an object and its image formed by a thin lens, (a) What is the focal length of the lens and what type of lens (converging or diverging) is it? (b) What is the height of the image? Is it real or virtual?

Figure 24.45

Problem 30.