Respuesta :
Answer:
The size of the image is 1.04 m.
Explanation:
Given that,
Height of object = 2.40 m
Distance of object = 2.60 m
Radius of curvature =4.00 m
Focal length [tex]f=\dfrac{R}{2}=\dfrac{4.00}{2}=2.00[/tex]
We need to calculate the image distance
Using mirror formula
[tex]\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{2.00}+\dfrac{1}{2.60}[/tex]
[tex]\dfrac{1}{v}= \dfrac{23}{26}[/tex]
[tex]v=1.13\ cm[/tex]
We need to calculate the height of the image
Using formula of magnification
[tex]m=\dfrac{h'}{h}=-\dfrac{v}{u}[/tex]
Put the value into the formula
[tex]\dfrac{h'}{2.40}=-\dfrac{1.13}{-2.60}[/tex]
[tex]h'=\dfrac{1.13}{2.60}\times2.40[/tex]
[tex]h'=1.04\ m[/tex]
Hence, The size of the image is 1.04 m
