A converging lens of
15cm focal length forms an image of an
object placed 9 cm from it. What is its
position and magnification?
To find the position of the image formed by the converging lens, we can use the lens equation:
1/f = 1/v - 1/u
Where:
- f is the focal length of the lens (15 cm)
- v is the distance of the image from the lens (unknown)
- u is the distance of the object from the lens (9 cm)
Substituting the given values into the equation:
1/15 = 1/v - 1/9
9v - 15u = uv
9v = uv + 15u
9v = u(v + 15)
v = u(v + 15)/9
Now, we can substitute the given values into the equation:
v = (9)(9 + 15)/9
v = (9)(24)/9
v = 24 cm
So, the distance of the image from the lens is 24 cm.
To find the magnification, we can use the formula:
magnification = -v/u
Using the given values:
magnification = -24/9
magnification = -8/3
So, the magnification of the image formed by the lens is -8/3.
To determine the position and magnification of the image formed by the converging lens, we can use the lens formula and magnification formula.
The lens formula is given by:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the distance of the image from the lens,
u is the distance of the object from the lens.
Given:
f = 15 cm
u = 9 cm
To find the position of the image (v), we can rearrange the lens formula:
1/v = 1/f - 1/u
Substituting the values:
1/v = 1/15 - 1/9
1/v = (3 - 5)/45
1/v = -2/45
v = -45/2 cm
However, since the image appears on the same side as the object (on the right side of the lens), the image distance should be positive. Therefore, we take the absolute value:
v = 45/2 cm
Now, to find the magnification (m), we use the magnification formula:
m = -v/u
Substituting the values:
m = -(45/2) / 9
m = -5/2
Therefore, the position of the image is 45/2 cm from the lens (on the right side) and the magnification is -5/2.