A 75 cm tall object is positioned 80 cm from a bi-concave lens with a focal length of 90 cm.

Question

**now i tried doing the problems, i just need feed back on if its correct or not. ****

1. calculate the distance where the image is formed
Used lens equation: (1/distance of object) + (1/distance of image) = 1/-focal length
The focal length is negative in the equation because it is a concave lens
(1/80cm) + (1/distance of image) = 1/-90 cm, the distance of the image equals -42.35

2. calculate the magnification of the object
magnification of object= -(distance of image/distance of object)
magnification of object = -(-42.35 cm/80 cm)
magnification of object = 0.529

3. calculate the size of the object
Magnification of object = (height of image/height of object)
0.529 = (height of image/75 cm)
height of image = 39.675

4. with this info we can say the image is virtual, upright, and diminished by magnification of 0.529x.

any feed back would be greatly appreciated.

Answer 1

Concave lenses always create virtual images
between the object and the lens

I agree with your work. I nearly always mess these up, so I have found first sketching a ray diagram helps me understand the math.