a diverging lens of focal length 30cm creates an image 3cm tall from an object that is 9cm tall where is the image located if the object is 60cm from the lens. what is the magnification

To solve this problem, we can use the lens formula:

1/f = 1/v - 1/u

Where:
- f is the focal length,
- v is the image distance from the lens,
- u is the object distance from the lens.

Given:
- f = 30 cm (focal length),
- u = -60 cm (since the object is placed 60 cm to the left of the lens; negative sign denotes distance on the left side of the lens).

First, let's calculate the value of v. Rearranging the lens formula, we have:

1/v = 1/f - 1/u

Substituting the given values, we get:

1/v = 1/30 - 1/(-60)
1/v = 1/30 + 1/60

To find the common denominator, we multiply both fractions by 2:

1/v = 2/60 + 1/60
1/v = 3/60

Now, let's invert both sides of the equation:

v = 60/3
v = 20 cm

So, the image is formed 20 cm away from the lens (positive sign denotes distance on the right side of the lens).

To find the magnification, we can use the formula:

magnification (m) = -v/u

Substituting the given values:

m = -(20 cm)/(-60 cm)
m = 1/3

Therefore, the magnification is 1/3.