A cup that is $6.0 \mathrm {~cm}$ tall is in front of a concave lens. The image of the cup is $2.5 \mathrm {~cm}$ tall and $7.5 \mathrm {~cm}$ from the lens. What is the focal length of the lens?
A) $- 13 \mathrm {~cm}$
B) $+ 6.0 \mathrm {~cm}$
C) $+ 18 \mathrm {~cm}$
D) $- 6.0 \mathrm {~cm}$
E) $- 7.5 \mathrm {~cm}$

Question

Quizplus · Accepted Answer

The magnification (m) of the lens is given by the ratio of the image height to the object height, $m = \frac{-2.5}{6.0}$. The negative sign indicates that the image is inverted. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, and knowing that magnification $m = -\frac{v}{u}$, we can solve for the focal length $f$. Given $m = -\frac{2.5}{6}$ and $v = -7.5$ cm (negative because the image formed by a concave lens is virtual and on the same side as the object), we find $f = -13$ cm.

A Cup That Is 6.0 cm6.0 \mathrm {~cm}6.0 cm Tall Is in Front of a Concave Lens

A Cup That Is $6.0 \mathrm {~cm}$ Tall Is in Front of a Concave Lens