Answer:
Explanation:
Given:
height of the object [tex]h_o=1.90m[/tex]
height of the image [tex]h_i=\frac{h_o}{2}\\=h_i=\frac{1.80}{2}=0.90m[/tex]
object distance [tex]d_o=3.50m[/tex]
we know that: [tex]\frac{h_i}{h_o}=\frac{-d_i}{d_o}\\\\=d_i=\frac{-h_i}{h_o}\times d_o\\\\=-\frac{0.90}{1.80}\times (3.50)=-1.75m[/tex]
image [tex]d_i=-1.75m[/tex]
According to lens formular:
[tex]\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}\\\\=\frac{1}{3.50}+-\frac{1}{1.75}\\\\f=-3.5[/tex]
focal length=[tex]\frac{radius}{2}\\\\radius=2\times f\\=2\times 3.5=7m[/tex]
