According to the question;
Focal length (f) = 12cm;
Object distance (u) = -8cm;
By mirror formula;
\(\frac1v\,+\,\frac1u\,=\,\frac1f\)
⇒ \(\frac1v\,+\,\frac1{-8}\,=\,\frac1{12}\)
⇒ \(\frac1v\,+\,\frac1{8}\,=\,\frac1{12}\)
\(\frac1v\,=\,\frac{3+2}{24}\)
⇒ \(\frac1v\,=\,\frac{5}{24}\)
⇒ v = \(\frac{5}{24}\) = 4.8 cm.
Since v = 4.8cm which is positive hence image is behind the mirror.
The image is formed at the distance of 4.8 cm behind the mirror and is virtual and erect.
