When a diverging lens is held 13.0 cm above a line of print, as in Figure 26.29, the image is 5.0 cm beneath the lens. (a) Is the image real or virtual? (b) What is the focal length of the lens?
When a diverging lens is held 13.0 cm above a line of print, as in Figure 26.29, the image is 5.0 cm beneath the lens. (a) Is the image real or virtual? (b) What is the focal length of the lens?
Homework Answers
Given
$$ \begin{array}{l} o=13.0 \mathrm{~cm} \\ i=5.0 \mathrm{~cm} \end{array} $$
(a)
The image formed is virtual
(b)
The focal length is
$$ \begin{array}{l} f=\frac{u v}{u+v} \\ =\frac{(13.0 \mathrm{~cm}) \times(-5.0 \mathrm{~cm})}{(13.0 \mathrm{~cm})+(-5.0 \mathrm{~cm})} \\ \hline f=-8.125 \mathrm{~cm} \end{array} $$
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When a diverging lens is held 13.0 cm above a line of print, as in Figure 26.29, the image is 5.0 cm beneath the lens. (a) Is the image real or virtual? (b) What is the focal length of the lens?
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A diverging lens has a focal length of ?12.0 cm. Real or virtual choices = real...
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Real or virtual choices = real or virtual
Orientation= upright or inverted
Relative Size= D(diminished) or E(enlarged)
Find the image distance for objects placed these distances from
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case, describe the image as real or virtual, upright or inverted,
and enlarged or diminished in size (select E for enlarged or D for
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