Optics and magnification problems

Question

1.An object is placed 0.22 m away from a concave lens and has a height of 0.08 m . The image produce is -0.03 m tall. Where is the image located?

2.A 0.38 m tall object is placed 0.43 m from a converging lens with a 0.19 m focal length. How tall is the image?

3.A concave mirror has a focal length of 0.50 m. If an object produces a virtual image 0.03 m from the mirror, where is it located?

Answer 1

1) The image is inverted we know the image is real so the magnification is 0.03/0.08 = -0.375. Since m=-di/do we can write -0.375=-di/0.22 so the image is located 0.0825 m (8.25 cms) in front of the mirror

2) Using 1/di+1/d0=1/f we know di =+0.34 m, the magnification is -0.34/0.43 = -0.79, so the image is 0.3 meters tall and is inverted and real

3) Using 1/do-1/0.03 = 1/0.5 we know do = 0.01875 (1.875 cms)