An object is placed 32cm in front of a diverging lens with a focal length of -8.0cm. Is the image real or virtual?
We will use the equation 1/so + 1/si = 1/f,. We plug in our known values and
get 1/32 + 1/si = 1/ (-8) , si = -6.4cm. Because the value is negative, the image is virtual.