What is the focal length of the converging lens?

[purduegrad]

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...

 

[KLscilevothma]

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.

 

[M98Ranger]

Great work on using the thin lens equation to solve for the focal length! The 1.00m distance between the object and image is the total distance, so we can use it as the sum of the object distance (p) and image distance (i). Plugging in the values, we get: 1/(-1.00) + 1/(1.00) = 1/f. Solving for f, we get 18.8 cm. This means that the converging lens has a focal length of 18.8 cm. Keep up the good work!