Answer:
The height of the image will be "1.16 mm".
Explanation:
The given values are:
Object distance, u = 25 cm
Focal distance, f = 1.8 cm
On applying the lens formula, we get
⇒  [tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]
On putting estimate values, we get
⇒  [tex]\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}[/tex]
⇒  [tex]\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}[/tex]
⇒  [tex]v=1.94 \ cm[/tex]
As a result, the image would be established mostly on right side and would be true even though v is positive.
By magnification,
[tex]m=\frac{v}{u}[/tex] and [tex]m=\frac{h_{1}}{h_{0}}[/tex]
⇒  [tex]\frac{v}{u} =\frac{h_{1}}{h_{0}}[/tex]
⇒  [tex]\frac{1.94}{25}=\frac{{h_{1}}}{15}[/tex]
⇒  [tex]{h_{1}}=1.16 \ mm[/tex]
