SP4. An object 4 cm tall is located 30 cm from a concave mirror with a focal length of 15 cm . Because the focal length is half the radius of the curvature, the object is located at the center of curvature of the mirror. a. Using the object-image distance formula, find the loca- tion of the image. b. Calculate the magnification of this image. c. Is the image real or virtual, upright or inverted? d. Trace two rays from the top of the object to confirm your results.

To find the location of the image, we start with the mirror formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Plugging in \( f = -15 \, \text{cm} \) (negative because it's a concave mirror) and \( d_o = -30 \, \text{cm} \), we get: \[ \frac{1}{-15} = \frac{1}{-30} + \frac{1}{d_i} \] Solving for \( d_i \): \[ \frac{1}{d_i} = \frac{1}{-15} + \frac{1}{30} \implies \frac{1}{d_i} = -\frac{2}{30} + \frac{1}{30} = -\frac{1}{30} \] So, \[ d_i = -30 \, \text{cm} \] This indicates the image is located 30 cm in front of the mirror. Next, we calculate magnification using the formula \( m = -\frac{d_i}{d_o} \): \[ m = -\frac{-30}{-30} = 1 \] The magnification of 1 means the image is the same size as the object. Now, evaluating if the image is real or virtual: since the image distance is negative and on the same side as the object in a concave mirror, the image is real and inverted. To visualize, trace two rays from the top of the object: one ray parallel to the principal axis reflects through the focal point, and another traveling through the focal point reflects parallel to the principal axis. Both rays converge to form the image, confirming it's real, inverted, and of the same height as the object!