Optics and images homework (1 Viewer)

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A real image is formed by a converging lens. It is three times higher than the object and inverted. the object and image are 1.00m apart. what is the focal length of the lens?

I assume 3 = the magnitude...the 1.00 m is confusing me since its the total distance

i tried using 1/p + 1/i = 1/f, to try to get the focal length....apparently its 18.8cm...
 
Let p be the distance between the object and the lens, then the distance between the image and the lens is 3p since the magnification is 3.

By applying
1/p + 1/i = 1/f

we have
[tex]\frac{1}{p} + \frac{1}{3p} = \frac{1}{f} [/tex] ......... (1)

Since the object and the image is 1m apart,
p + 3p = 1 ....................(2)

By solving (1) and (2), you'll get 18.75 cm.
 

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