Respuesta :
Explanation:
It is given that,
Height of the object, h = 3 cm
Object distance, u = -45 cm
Focal length of the mirror, f = - 25 cm
Using mirror's formula as :
[tex]\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}[/tex], v is the image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{-25}-\dfrac{1}{-45}[/tex]
v = −56.25 cm
Let h' is the height of the image. Using the formula of magnification as :
[tex]m=\dfrac{-v}{u}=\dfrac{h'}{h}[/tex]
[tex]h'=\dfrac{-vh}{u}[/tex]
[tex]h'=\dfrac{-(-56.25)\times 3}{-45}[/tex]
[tex]h'=+3.75\ cm[/tex]
So, the image is larger than object. The image is upright and inverted. Hence, this is the required solution.
