Question 23.7: Images Formed by a Converging Lens Goal Calculate geometric ...
Images Formed by a Converging Lens
Goal Calculate geometric quantities associated with a converging lens.
Problem A converging lens of focal length 10.0 \mathrm{~cm} forms images of an object situated at various distances. (a) If the object is placed 30.0 \mathrm{~cm} from the lens, locate the image, state whether it’s real or virtual, and find its magnification. (b) Repeat the problem when the object is at 10.0 \mathrm{~cm} and (c) again when the object is 5.00 \mathrm{~cm} from the lens.
Strategy All three problems require only substitution into the thin-lens equation and the associated magnification equation-Equations 23.10 and 23.11,
M={\frac{h^{\prime}}{h}}=\,-\,{\frac{q}{\rho}} (23.10)
{\frac{1}{p}}+{\frac{1}{q}}={\frac{1}{f}} (23.11)
respectively. The conventions of Table 23.3 must be followed.
TABLE 23.3
Sign Conventions for Thin Lenses
| Quantity | Symbol | In Front | In Back | Convergent | Divergent |
| Object location | p | + | — | ||
| Image location | q | — | + | ||
| Lens Radii Focal | R 1, R 2 | — | + | ||
| Length | f | + | — |
(a) Find the image distance and describe the image when the object is placed at 30.0 \mathrm{~cm}.
The ray diagram is shown in Figure 23.26a. Substitute into the thin-lens equation to locate the image:
\begin{gathered} \frac{1}{p}+\frac{1}{q}=\frac{1}{f} \\ \frac{1}{30.0 \mathrm{~cm}}+\frac{1}{q}=\frac{1}{10.0 \mathrm{~cm}} \end{gathered}
Solve for q, the image distance. It’s positive, so the image is real and on the far side of the lens.
q=+15.0 \mathrm{~cm}
The magnification of the lens is obtained from Equation 23.10. M is negative and less than one in absolute value, so the image is inverted and smaller than the object:
M=-\frac{q}{p}=-\frac{15.0 \mathrm{~cm}}{30.0 \mathrm{~cm}}=-0.500
(b) Repeat the problem, when the object is placed at 10.0 \mathrm{~cm}.
Locate the image by substituting into the thin-lens equation:
\frac{1}{10.0 \mathrm{~cm}}+\frac{1}{q}=\frac{1}{10.0 \mathrm{~cm}} \rightarrow \frac{1}{q}=0
This equation is satisfied only in the limit as q becomes infinite.
q \rightarrow \infty
(c) Repeat the problem when the object is placed 5.00 \mathrm{~cm} from the lens.
See the ray diagram in Figure 23.26b. Substitute into the thin-lens equation to locate the image:
\frac{1}{5.00 \mathrm{~cm}}+\frac{1}{q}=\frac{1}{10.0 \mathrm{~cm}}
Solve for q, which is negative, meaning the image is on the same side as the object and is virtual:
q=-10.0 \mathrm{~cm}
Substitute the values of p and q into the magnification equation. M is positive and larger than one, so the image is upright and double the object size:
M=-\frac{q}{p}=-\left(\frac{-10.0 \mathrm{~cm}}{5.00 \mathrm{~cm}}\right)=+2.00
Remarks The ability of a lens to magnify objects led to the inventions of reading glasses, microscopes, and telescopes.
