Answer
(a) $12cm$ to the right of the lens
(b) virtual
(c) $0.174$
(d) upright
Work Step by Step
(a) We know that
$\frac{1}{f}=\frac{1}{d_{\circ}}+\frac{1}{d_i}$
$\implies \frac{1}{-20cm}=\frac{1}{25cm}+\frac{1}{d_i}$
$\implies d_i=-11.11cm$
The image distance from the mirror is $11.11cm+x=11.11cm+10cm=21.11cm$
Now this becomes the object at a distance of $21.11cm+10cm=31.11cm$ behind the lens
$\frac{1}{f}=\frac{1}{d_{\circ}^{\prime}}+\frac{1}{di^{\prime}}$
$\implies \frac{-1}{20cm}=\frac{1}{31.11cm}+\frac{1}{d_i^{\prime}}$
$\implies d_i^{\prime}=12cm$ behind the mirror or the lens.
(b) Since we know that the light does not pass through the image, therefore, it is a virtual image.
(c) We know that the magnification is given as
$M=(\frac{d_i}{d_{\circ}})(\frac{di^{\prime}}{d_{\circ}^{\prime}})$
We plug in the known values to obtain:
$M=(\frac{-11.11cm}{25cm})(\frac{-12.17cm}{31.11cm})$
$\implies M=0.174$
(d) We know that the magnification is greater than zero, and therefore, the image is upright.
