91Ó°ÊÓ

When dealing with convex mirrors, we use the negative magnification. Thus, the given magnification in this exercise is \(-0.60\), which can be rewritten as a decimal \((-0.60) \times = (-0.60)\). The magnification formula is defined as: $$ m = \frac{h_{i}}{h_{o}} = \frac{d_{i}}{d_{o}} $$ Where \(m\) is the magnification, \(h_{i}\) is the image height, \(h_{o}\) is the object height, \(d_{i}\) is the image distance, and \(d_{o}\) is the object distance. We are given \(m=-0.60\) and \(d_{o}=2.0\mathrm{~m}\). We can solve for the image distance, \(d_{i}\): $$ d_{i} = m \times d_{o} = -0.60 \times 2.0\mathrm{~m} $$

We calculate the image distance: $$ d_{i} = -0.60 \times 2.0\mathrm{~m} = -1.2\mathrm{~m} $$ Now we apply the mirror formula for a convex mirror, which is: $$ \frac{1}{f} = \frac{1}{d_{o}} - \frac{1}{d_{i}} $$ Where \(f\) is the focal length. We have \(d_{o}=2.0\mathrm{~m}\) and \(d_{i} = -1.2\mathrm{~m}\). Plugging in these values, we get: $$ \frac{1}{f} = \frac{1}{2.0\mathrm{~m}} - \frac{1}{-1.2\mathrm{~m}} $$ Now we can find the common denominator and solve for \(f\): $$ \frac{1}{f} = \frac{-1.2 \mathrm{~m} + 2.0 \mathrm{~m}}{2.0\mathrm{~m} \times -1.2\mathrm{~m}} = \frac{0.8\mathrm{~m}}{-2.4\mathrm{~m^2}} $$ Multiplying both sides by \(f\) and \((-2.4\mathrm{~m}^2)\), we obtain: $$ f = \frac{-2.4\mathrm{~m^2}}{0.8\mathrm{~m}} $$ $$ f = -3\mathrm{~m} $$ The focal length of the convex mirror is \(-3.0\mathrm{~m}\). The negative sign indicates that the focal length is on the same side as the object, which is a characteristic of convex mirrors.