Answer:
Data given.
focal length (f)=15m÷2=7.5m
Distance of the object(U)=10m
Image distance (v)=?
Magnification (M)=?
Solution:
From:
1/f=1/u+1/v
1/7.5=1/10+1/v=75
then v=75m
Magnification, M=u/v
=75/10=7.5
Then magnification=7.5
Answer:
v = 30 m and m = 3
Explanation:
Given that,
The radius of curvature of the mirror, R = 15 m
Focal length, f = 7.5 m
Object distance, u = -10 m
We need to find the image distance and the magnification of the object.
Using mirror's formula,
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(7.5)}+\dfrac{1}{(-10)}\\\\v=30\ m[/tex]
The magnification of the object in mirror is given by :
[tex]m=\dfrac{-v}{u}\\\\m=\dfrac{-30}{-10}\\\\m=3[/tex]
So, the distance of the image from the mirror and the magnification of the object are 30 m and 3 respectively.
