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$\cdot$ You want to view an insect 2.00 $\mathrm{mm}$ in length through a magnifier. If the insect is to be at the focal point of the magnifier, what focal length will give the image of the insect an angular size of 0.025 radian?
$\begin{aligned} \text { REFLECT: } & \\ \text { The angular size of the insect when it is at the near point of a nomal eye is } \\ \theta &=\frac{y}{25 \mathrm{cm}} \\ &=\frac{2.0 \mathrm{mm}}{250 \mathrm{mm}} \\ &=0.0080 \mathrm{rad} . \end{aligned}$
Physics 102 Electricity and Magnetism
Physics 103
Chapter 25
Optical Instruments
Electromagnetic Waves
Reflection and Refraction of Light
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okay. So angular size of an object. His ah data is given by length over focal in't. Why over f and so were given why we're given data. So if the focal length of the object is length of regular size and length is two millimeters, that's the length of the infect. Over data is 0.25 Radiance is the right unit, and so that works out to be 80 millimeters are a centimeters for the focal length. Um, for the focal length off, no of the object. And this will in fact be the focal length off times. This will in fact be the focal length of the image. Because we are because the the insect is at the focal point. And we know that negative s prime over us is equal to y prime over. Why? So that so That why prime over s prime is equal to why, over us of or at least a magnitude czar. And so the height of the image. And so the height of the image, the ratio of the height, two of the image to the length of the image. To them, image distance is equal to ratio of the length of the object to your object distance on DH. So the and so fade off will remain the same. And so well, why?
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