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You can argue that a flat piece of glass, such as in a window, is like a lens with an infinite focal length. If so, where does it form an image? That is, how are ${d}_{\text{i}}$ and ${d}_{\text{o}}$ related?
When you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be a fixed distance from the film for both near and distant objects?
The focal length of the lens is fixed, so the image distance changes as a function of object distance.
A thin lens has two focal points, one on either side of the lens at equal distances from its center, and should behave the same for light entering from either side. Look backward and forward through a pair of eyeglasses and comment on whether they are thin lenses.
Will the focal length of a lens change when it is submerged in water? Explain.
Yes, the focal length will change. The lens maker’s equation shows that the focal length depends on the index of refraction of the medium surrounding the lens. Because the index of refraction of water differs from that of air, the focal length of the lens will change when submerged in water.
How far from the lens must the film in a camera be, if the lens has a 35.0-mm focal length and is being used to photograph a flower 75.0 cm away? Explicitly show how you follow the steps in the [link] .
A certain slide projector has a 100 mm-focal length lens. (a) How far away is the screen if a slide is placed 103 mm from the lens and produces a sharp image? (b) If the slide is 24.0 by 36.0 mm, what are the dimensions of the image? Explicitly show how you follow the steps in the [link] .
a.
$\frac{1}{{d}_{\text{i}}}+\frac{1}{{d}_{\text{o}}}=\frac{1}{f}\Rightarrow {d}_{\text{i}}=3.43\phantom{\rule{0.2em}{0ex}}\text{m}$ ;
b.
$m=\mathrm{-33.33}$ , so that
$\begin{array}{c}\left(2.40\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}\phantom{\rule{0.2em}{0ex}}\text{m}\right)(33.33)=80.0\phantom{\rule{0.2em}{0ex}}\text{cm, and}\hfill \\ \left(3.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}\phantom{\rule{0.2em}{0ex}}\text{m}\right)(33.33)=1.20\phantom{\rule{0.2em}{0ex}}\text{m}\Rightarrow 0.800\phantom{\rule{0.2em}{0ex}}\text{m}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}1.20\phantom{\rule{0.2em}{0ex}}\text{m or}\phantom{\rule{0.2em}{0ex}}80.0\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}120\phantom{\rule{0.2em}{0ex}}\text{cm}\hfill \end{array}$
A doctor examines a mole with a 15.0-cm focal length magnifying glass held 13.5 cm from the mole. (a) Where is the image? (b) What is its magnification? (c) How big is the image of a 5.00 mm diameter mole?
A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens must the film be? (b) If the film is 36.0 mm high, what fraction of a 1.75-m-tall person will fit on it? (c) Discuss how reasonable this seems, based on your experience in taking or posing for photographs.
a.
$\begin{array}{c}\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{f}\hfill \\ {d}_{\text{i}}=5.08\phantom{\rule{0.2em}{0ex}}\text{cm}\hfill \end{array}$ ;
b.
$m=\mathrm{-1.695}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}$ , so the maximum height is
$\frac{0.036\phantom{\rule{0.2em}{0ex}}\text{m}}{1.695\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}}=2.12\phantom{\rule{0.2em}{0ex}}\text{m}\Rightarrow \text{100\%}\phantom{\rule{0.2em}{0ex}}$ ;
c. This seems quite reasonable, since at 3.00 m it is possible to get a full length picture of a person.
A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is 33.0 mm. (a) What is the closest object that can be photographed? (b) What is the magnification of this closest object?
Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus? (b) What is the height of the object if its image is 2.00 cm high?
a.
$\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{f}\Rightarrow {d}_{\text{o}}=2.55\phantom{\rule{0.2em}{0ex}}\text{m}$ ;
b.
$\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=\text{\u2212}\frac{{d}_{\text{i}}}{{d}_{\text{o}}}\Rightarrow {h}_{\text{o}}=1.00\phantom{\rule{0.2em}{0ex}}\text{m}$
What is the focal length of a magnifying glass that produces a magnification of 3.00 when held 5.00 cm from an object, such as a rare coin?
The magnification of a book held 7.50 cm from a 10.0 cm-focal length lens is 3.00. (a) Find the magnification for the book when it is held 8.50 cm from the magnifier. (b) Repeat for the book held 9.50 cm from the magnifier. (c) Comment on how magnification changes as the object distance increases as in these two calculations.
a. Using $\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{f}$ , ${d}_{\text{i}}=\mathrm{-56.67}\phantom{\rule{0.2em}{0ex}}\text{cm}$ . Then we can determine the magnification, $m=6.67$ . b. ${d}_{\text{i}}=\mathrm{-190}\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}$ and $m=+20.0$ ; c. The magnification m increases rapidly as you increase the object distance toward the focal length.
Suppose a 200 mm-focal length telephoto lens is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1000 m high cliff on one of the mountains?
A camera with a 100 mm-focal length lens is used to photograph the sun. What is the height of the image of the sun on the film, given the sun is $1.40\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{km}$ in diameter and is $1.50\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{km}$ away?
$\begin{array}{cc}\hfill \frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}& =\frac{1}{f}\hfill \\ \hfill {d}_{\text{i}}& =\frac{1}{(1\text{/}f)-(1\text{/}{d}_{\text{o}})}\hfill \\ \hfill \frac{{d}_{\text{i}}}{{d}_{\text{o}}}& =6.667\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-13}}=\frac{{h}_{\text{i}}}{{h}_{\text{o}}}\hfill \\ \hfill {h}_{\text{i}}& =\mathrm{-0.933}\phantom{\rule{0.2em}{0ex}}\text{mm}\hfill \end{array}$
Use the thin-lens equation to show that the magnification for a thin lens is determined by its focal length and the object distance and is given by $m=f\text{/}(f-{d}_{\text{o}})$ .
An object of height 3.0 cm is placed 5.0 cm in front of a converging lens of focal length 20 cm and observed from the other side. Where and how large is the image?
${d}_{\text{i}}=\mathrm{-6.7}\phantom{\rule{0.2em}{0ex}}\text{cm}$
${h}_{\text{i}}=4.0\phantom{\rule{0.2em}{0ex}}\text{cm}$
An object of height 3.0 cm is placed at 5.0 cm in front of a diverging lens of focal length 20 cm and observed from the other side. Where and how large is the image?
An object of height 3.0 cm is placed at 25 cm in front of a diverging lens of focal length 20 cm. Behind the diverging lens, there is a converging lens of focal length 20 cm. The distance between the lenses is 5.0 cm. Find the location and size of the final image.
83 cm to the right of the converging lens, $m=\mathrm{-2.3},{h}_{\text{i}}=6.9\phantom{\rule{0.2em}{0ex}}\text{cm}$
Two convex lenses of focal lengths 20 cm and 10 cm are placed 30 cm apart, with the lens with the longer focal length on the right. An object of height 2.0 cm is placed midway between them and observed through each lens from the left and from the right. Describe what you will see, such as where the image(s) will appear, whether they will be upright or inverted and their magnifications.
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