ANSWER:
The mirror’s magnification is 0.06
The focal length is equal to 19.15 cm.
The mirror's radius of curvature is 38.3 cm
Explanation:
Given:
height image = 9 cm
height object = 1.5 m = 150 cm
The mirror's magnification is calculated using the following formula:
![m=(h_i)/(h_o)=(9)/(150)=0.06](https://img.qammunity.org/2023/formulas/physics/college/ovagtid0sif8d91px182681cd6i8vgmbdk.png)
The mirror’s magnification is 0.06
Since we know the distance of the object which is -3 m (-300 cm) and the mirror's magnification we can calculate the distance of the image, just like this:
![\begin{gathered} m=(-d_i)/(d_o) \\ \\ 0.06=(-d_i)/(-300) \\ \\ d_i=300\cdot0.06 \\ \\ d_i=18\text{ cm} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/9rqjxczmq7e21vlzpemw2t6cdlv36qhbfe.png)
We calculate the focal length using the following formula:
![\begin{gathered} (1)/(f)=(1)/(d_i)+(1)/(d_o) \\ \\ (1)/(f)=(1)/(18)+(1)/(-300) \\ \\ (1)/(f)=(18-300)/(-5400) \\ \\ f=(-5400)/(-282) \\ \\ f=\:19.15\text{ cm} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/g3ak89arukg5h4ubi8ewnnjcnbcnmahbvq.png)
The focal length is equal to 19.15 cm.
Finally we calculate the mirror's radius of curvature knowing that twice the focal length, therefore:
![\begin{gathered} r=2f=19.15\cdot2 \\ \\ r=38.3\text{ cm} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/fa7c8vht1ow8kkpamn3gz1rc3tafynhoxt.png)
The mirror's radius of curvature is 38.3 cm