Question
A diverging lens has a focal length of $15.0 \mathrm{cm} .$ An object placed near it forms a 2.0 -cm-high image at a distance of $5.0 \mathrm{cm}$ from the lens.a. What are the object position and object height?b. The diverging lens is now replaced by a converging lens with the same focal length. What are the image position, height, and orientation? Is it a virtual image or a real image?
Step 1
For a diverging lens, the focal length is negative, so $f = -15.0 \, \text{cm}$. The image distance is also given as $d_i = -5.0 \, \text{cm}$. We can substitute these values into the lens formula to find the object distance. Show more…
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