Answer:
The size of the image is 1.04 m.
Step-by-step explanation:
Given that,
Height of object = 2.40 m
Distance of object = 2.60 m
Radius of curvature =4.00 m
Focal length
![f=(R)/(2)=(4.00)/(2)=2.00](https://img.qammunity.org/2020/formulas/physics/college/z4ge23ron7jzftlrv0a9zwb6hm03y3he0i.png)
We need to calculate the image distance
Using mirror formula
![(1)/(f)=(1)/(v)+(1)/(u)](https://img.qammunity.org/2020/formulas/physics/college/94otoa71va28a3hmd2g5tfkcdbdipqdm7r.png)
![(1)/(v)=(1)/(2.00)+(1)/(2.60)](https://img.qammunity.org/2020/formulas/physics/college/4dqbufh4yb5ogyl1dt6vn1v2u8ktusmypx.png)
![(1)/(v)= (23)/(26)](https://img.qammunity.org/2020/formulas/physics/college/n7yi8aj2l1r3172iw5nesbabuf0wthwc6v.png)
![v=1.13\ cm](https://img.qammunity.org/2020/formulas/physics/college/cbe8i3igfo5qv9boy4efh3fcv5yexfvj36.png)
We need to calculate the height of the image
Using formula of magnification
![m=(h')/(h)=-(v)/(u)](https://img.qammunity.org/2020/formulas/physics/college/fhlxv13k5f1kpvgklkvfoju3eqd187e77d.png)
Put the value into the formula
![(h')/(2.40)=-(1.13)/(-2.60)](https://img.qammunity.org/2020/formulas/physics/college/ekz3vo84114s91qkrvbbyjom8uejp8zpgg.png)
![h'=(1.13)/(2.60)*2.40](https://img.qammunity.org/2020/formulas/physics/college/fbo7knt4bficqrcba9nk90socbjosw6g1s.png)
![h'=1.04\ m](https://img.qammunity.org/2020/formulas/physics/college/1gg4o9yll8z5q5s9zeixw6sxkv1xrytu33.png)
Hence, The size of the image is 1.04 m