Chapter 24, Problem 34P

Set Up: Use $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ to calculate f and then use $\begin{array}{l}\\ \frac{1}{s}+\frac{1}{s\prime }=\frac{1}{f}\end{array}$ to locate the image, s = 18.0 cm.

Solve: (a) $\frac{1}{f}=(0.5)\left(\frac{1}{10.0\text{ cm}}-\frac{1}{-15.0\text{ cm}}\right)$and f = +12.0 cm. $\frac{1}{s\prime }=\frac{1}{f}-\frac{1}{s}=\frac{s-f}{sf}$.

$s\prime =\frac{f}{s-f}=\frac{(18.0\text{ cm})(12.0\text{ cm})}{18.0\text{ cm}-12.0\text{ cm}}=+36.0\text{ cm}$.

The image is 36.0 cm to the right of the lens.

(b) $\frac{1}{f}=(0.5)\left(\frac{1}{10.0\text{ cm}}-\frac{1}{\infty }\right)$ and f = +20.0 cm

$s\prime =\frac{sf}{s-f}=\frac{(18.0\text{ cm})(20.0\text{ cm})}{18.0\text{ cm}-20.0\text{ cm}}=-180\text{ cm}$

The image is 180 cm to the right of the lens.

(c) $\frac{1}{f}=(0.5)\left(\frac{1}{-15.0\text{ cm}}-\frac{1}{10.0\text{ cm}}\right)$ and f = −12.0 cm

$s\prime =\frac{sf}{s-f}=\frac{(18.0\text{ cm})(-12.0\text{ cm})}{18.0\text{ cm}+12.0\text{ cm}}=-7.2\text{ cm}$

The image is 7.2 cm to the right of the lens.

(d) $\frac{1}{f}=(0.5)\left(\frac{1}{-10.0\text{ cm}}-\frac{1}{-15.0\text{ cm}}\right)$ and f = −60.0 cm

$s\prime =\frac{sf}{s-f}=\frac{(18.0\text{ cm})(-60.0\text{ cm})}{18.0\text{ cm}+60.0\text{ cm}}=-13.8\text{ cm}$

The image is 13.8 cm to the right of the lens.

34. For each thin lens shown in Figure 24.47, calculate the location of the image of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes.

Figure 24.47

Problem 34