The object (upright arrow) in the following system has a height of \(2.5 \mathrm{~cm}\) and is placed \(5.0 \mathrm{~cm}\) away from a converging (convex) lens with a focal length of \(3.0 \mathrm{~cm}\). What is the magnification of the image? Is the image upright or inverted? Confirm your answers by ray tracing.

To find the image distance, we can use the lens formula which relates object distance (denoted as \(d_o\)), image distance (denoted as \(d_i\)), and focal length (denoted as \(f\)): $$ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $$ Given that the object distance \(d_o = 5.0\mathrm{~cm}\) and the focal length \(f = 3.0\mathrm{~cm}\), we can plug in the values and solve for the image distance \(d_i\): $$ \frac{1}{3} = \frac{1}{5} + \frac{1}{d_i} $$ $$ \frac{1}{d_i} = \frac{1}{3} - \frac{1}{5} $$ $$ d_i = \frac{15}{2} \mathrm{~cm} $$

Now that we have the image distance, we can use the magnification formula to find the magnification (denoted as \(M\)): $$ M = - \frac{d_i}{d_o} $$ Plugging in the values, we get: $$ M = - \frac{15/2}{5} $$ $$ M = - \frac{3}{2} $$ The magnification is \(-1.5\).

Since the magnification is negative, the image is inverted (upside down). To confirm this using ray tracing, we can draw three rays: 1. A ray parallel to the principal axis from the object, which goes through the focal point after passing through the lens. 2. A ray that goes through the center of the lens, which continues straight without any bending. 3. A ray that goes through the focal point on the same side as the object and becomes parallel to the principal axis after passing through the lens. These rays will converge at a point, showing the location of the image. If we draw these rays according to the given problem, we will find that the image formed is inverted (upside down) which confirms our result from the magnification calculation.