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

**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.

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.

Your calculations for the distance where the image is formed and the magnification of the object are correct. However, there seems to be a mistake in your calculation for the size of the object.

Let's revisit the calculation for the size of the object:

The magnification of the object is given by:

Magnification of object = (height of image / height of object)

You have correctly calculated the magnification of the object to be 0.529.

To find the height of the image, we can use the relationship:

Magnification of object = (height of image / height of object)

0.529 = (height of image / 75 cm)

Rearranging the equation, we find:

height of image = (0.529) * 75 cm

height of image = 39.675 cm

So the correct size of the object should be 39.675 cm, not 39.675 cm/s.

Other than that, your analysis of the image being virtual, upright, and diminished is correct. Keep up the good work!