Answer
(a) $16cm$
(b) real
(c) $-0.80cm$
(d) inverted
Work Step by Step
(a) We know that
$\frac{1}{f}=\frac{1}{d_{\circ}^{\prime}}+\frac{1}{d_{i}^{\prime}}$
$\frac{1}{20cm}=-\frac{1}{80cm}+\frac{1}{d_i^{\prime}}$
This simplifies to:
$d_i^{\prime}=16cm$ in front of the lens.
(b) We know that since the light passes through the image, therefore, the image is a real image.
(c) We know that
$M=(\frac{d_i}{d_{\circ}})(\frac{d_i^{\prime}}{d_{\circ}^{\prime}})$
We plug in the known values to obtain:
$M=(\frac{100cm}{25cm})(\frac{16cm}{-80cm})$
$\implies M=-0.80cm$
(d) We know that as the magnification is less than zero; therefore, the image is inverted.
