Question

when using the lens equation, a negative value as the solution for d, indicates that the image is
real.
virtual.
upright.
inverted.

Explanation:

In optics, when using the lens equation \(\frac{1}{f}=\frac{1}{d_{o}}+\frac{1}{d_{i}}\) (where \(f\) is the focal length, \(d_{o}\) is the object distance, and \(d_{i}\) is the image distance), a negative value of \(d_{i}\) indicates that the image is virtual. A real image has a positive \(d_{i}\) (formed on the opposite side of the lens from the object for a converging lens in normal imaging situations). The terms "upright" and "inverted" are related to the magnification (\(m =-\frac{d_{i}}{d_{o}}\)) rather than directly to the sign of \(d_{i}\) in terms of classifying an image as real or virtual. So, a negative \(d_{i}\) means the image is virtual.

Answer:

virtual.