Answer:
Height of the image is 4.79 cm.
Explanation:
Object distance, u = -11 cm
Focal length of the mirror, f = -24 cm
Height of the object, h = 2.6 cm
We need to find the height of the image. Firstly, using the mirror's formula as :
[tex]\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}[/tex]
v is the image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{(-24)}-\dfrac{1}{(-11)}[/tex]
v = 20.30 cm
The magnification of the image is given by :
[tex]m=\dfrac{-v}{u}=\dfrac{h'}{h}[/tex], h' is the height of the image
[tex]\dfrac{-v}{u}\times h={h'}[/tex]
[tex]\dfrac{-20.30}{-11}\times 2.6={h'}[/tex]
h' = 4.79 cm
So, the height of the image is 4.79 cm. Hence, this is the required solution.
