According to the question;
Object distance (u) = -27cm;
Focal length (f) = -18cm;
Image distance = v;
By mirror formula;
\(\frac1v\,+\,\frac1u\,=\frac1f\)
⇒\(\frac1v\,+\,\frac{1}{-27}\,=\frac{1}{-18}\)
⇒ \(\frac1v\,=\,\frac1{27}\,-\,\frac1{18}\)
⇒\(\frac1v\,=\,\frac{-3+2}{54}\)
⇒\(\frac1v\,=\,-\frac{1}{54}\)
⇒ v = -54cm.
Thus, screen should be placed 54cm in front of the mirror to obtain the sharp focused image.
Height of object h1= 7cm;
Magnification =\(\frac{h_2}{h_1}\) = \(-\frac{v}{u}\)
Putting values of v and u
Magnification = \(\frac{h_2}{7}\) = \(-\frac{-54}{-27}\)
⇒\(\frac{h_2}{7}\) = -2
⇒ h2 = 7 ×-2 = -14.
Height of image is 14 cm.
Negative sign means image is real and inverted.
Thus real, inverted image of 14cm size is formed.