Question

converging, lens of 15crn
focal length forms an image of an
object placed 9cm from it. What is
its position and magnification?
A. 6cm and 3/2
B. 6cm and 1/3
C. 6cm and -2/3
D. 6cm and 2/3

Answer 1

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

1/f = 1/v - 1/u

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

Given:
- f = 15 cm
- u = 9 cm

We can rearrange the lens formula to solve for v:

1/v = 1/f + 1/u
1/v = 1/15 + 1/9
1/v = (3 + 5)/45
1/v = 8/45

Now, solving for v:

v = 45/8 cm

The magnification (M) is given by:

M = -v/u

Substituting the given values:

M = -(45/8) / 9
M = -45 / (8 * 9)
M = -5 / (8 * 3)
M = -5 / 24

Therefore, the position of the image is 45/8 cm and the magnification is -5/24.

None of the given options match the calculated values, so it seems there may be a mistake in the problem statement or the options.