According to the question;
Object distance = u;
Image distance (v) = -15cm;
Focal length = -20cm
By lens formula;
\(\frac1v-\frac1u=\frac1f\)
⇒ \(\frac1{15}-\frac1u=\frac1{-20}\)
⇒ \(\frac1{u}=\frac1{20}-\frac1{15}\)
⇒ \(\frac1{u}=\frac{3-4}{60}=\frac{-1}{60}\)
⇒ u = -60cm.
Therefore, object is placed at 60 cm in front of lens.
Now;
Height of object h1= 5cm;
Magnification = \(\frac{h_2}{h_1}=\frac{v}{u}\)
Putting values of v and u
Magnification = \(\frac{h_2}{5}=\frac{-15}{-60}\)
⇒ \(\frac{h_2}{5}=\frac{1}{4};\)
⇒ h2 = \(\frac54\) = 1.25
Height of image is 1.25 cm.
Positive sign means image is virtual.