Step-by-step 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 :
, v is the image distance
![(1)/(v)=(1)/(f)-(1)/(u)](https://img.qammunity.org/2020/formulas/physics/college/8syfwgrfiff7l3fejesh6gnv25gvk6h87v.png)
![(1)/(v)=(1)/(-25)-(1)/(-45)](https://img.qammunity.org/2020/formulas/physics/college/kibgnthmidtoealsukz2ssuyiedvf9nze1.png)
v = −56.25 cm
Let h' is the height of the image. Using the formula of magnification as :
![m=(-v)/(u)=(h')/(h)](https://img.qammunity.org/2020/formulas/physics/college/7rauqt2l0s1ezjyziwgygkx4l8g0wnvp7h.png)
![h'=(-vh)/(u)](https://img.qammunity.org/2020/formulas/physics/college/6i97yvwnhxtiuyfaspf6l76a6ladknoko1.png)
![h'=(-(-56.25)* 3)/(-45)](https://img.qammunity.org/2020/formulas/physics/college/ngq46tbtcfe3i9vj3h6shig7jq4f7mymew.png)
![h'=+3.75\ cm](https://img.qammunity.org/2020/formulas/physics/college/g3w2ak55i0gjj3crovejkme07w452rj0ui.png)
So, the image is larger than object. The image is upright and inverted. Hence, this is the required solution.