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
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.
To find the position and magnification of an image formed by a converging lens, we can use the lens formula and the 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 image distance, and
u is the object distance.
Given:
f = 15 cm (focal length),
u = 9 cm (object distance).
Let's substitute these values into the lens formula and solve for v:
1/15 = 1/v - 1/9
Multiply both sides of the equation by 135v to remove fractions:
9v - 15v = 15 * 9
-6v = 135
Divide both sides of the equation by -6:
v = -135/6
v = -22.5 cm
The negative sign indicates that the image formed is virtual, meaning it is on the same side as the object.
Now, let's find the magnification (M):
M = -v/u
Substitute the values for v and u:
M = -(-22.5)/9
M = 22.5/9
M = 2.5/1
M = 5/2
The magnification is 5/2 or 2.5.
Therefore, the position of the image is 22.5 cm (or -22.5 cm) and the magnification is 2.5. None of the given options (A, B, C, or D) match the calculated values.