Physics

A diverging lens always produces a virtual image on the same side as the object—and the lens equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ works perfectly here if you use the sign convention carefully. For a diverging lens, the focal length is negative, and the virtual image distance is also negative. Plug in your object and image distances with the right signs, and you'll solve directly for $f$.

A diverging lens always produces a virtual image on the same side as the object—and the lens equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ works perfectly here if you use the sign convention carefully. For a diverging lens, the focal length is negative, and the virtual image distance is also negative. Plug in your object and image distances with the right signs, and you'll solve directly for $f$.