Answer
(a) The image is located 8.3 cm from the same side as the virtual object.
(b) real
Work Step by Step
(a) We know that
$d_i=(\frac{1}{f}-\frac{1}{d_{\circ}})^{-1}$
$\implies d_i=(Refractive \space power-\frac{1}{d_{\circ}})^{-1}$
We plug in the known values to obtain:
$d_i=(3.75-\frac{1}{-0.12})^{-1}=8.3 cm$
The image is located 8.3 cm from the same side as the virtual object.
(b) As the light passes through the image, the image is real.
