Chapter 24, Problem 42P

Set Up: Apply $\frac{1}{s}+\frac{1}{s\prime }=\frac{1}{f}$ with f = ±35.0 cm. We know that the magnification is $m=-\frac{s\prime }{s}$

Solve: (a) We want the size of the image to be twice that of the object and so we must have m = ±2. Since the image is real we know that s′ > 0, which implies that m = −2 = $-\frac{s\prime }{s}$. Thus we conclude that s′ = 2s. Now we can determine the location of the object: $\frac{1}{s}+\frac{1}{s\prime }=\frac{1}{s}+\frac{1}{2s}=\frac{3}{2s}=\frac{1}{f}$. Solving for s we obtain $s=\frac{3}{2}f$. Since we know that s > 0 we must have that f = +35.0 cm. and thus $s=\frac{3}{2}f=\frac{3}{2}(35.0\text{ cm})=52.5\text{ cm}$.

(b) We again want the image to be twice the size as the object: however, in this case we have a virtual image so s′ < 0 and m = +2 = $-\frac{s\prime }{s}$. Thus, we have s′ = −2s. Now we can determine the location of the object: $\frac{1}{s}+\frac{1}{s\prime }=\frac{1}{s}+\frac{1}{-2s}=\frac{1}{2s}=\frac{1}{f}$. Solving for s we obtain $s=\frac{1}{2}f$. Since we know that s > 0 we must have that f = +35.0 cm and thus = $s=\frac{1}{2}(35.0\text{ cm})$ = 17.5 cm.

42. (a) You want to use a lens with a focal length of 35.0 cm to produce a real image of an object, with the image twice as long as the object itself. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification—what kind of lens do you need, and where should the object be placed?