Two converging lenses, each of focal length 15.1 cm, are placed 40.2 cm apart, and an object is placed 30.0 cm in front of the first lens. Where is the final image formed?
The image is located...
What is the magnification of the system?
M =
To determine the location of the final image formed by the two converging lenses, we can use the lens formula.
The lens formula states:
1/f = 1/v - 1/u
where:
f = focal length of the lens
v = distance of the image from the lens (positive if the image is on the same side as the observer, negative if the image is on the opposite side)
u = distance of the object from the lens (positive if the object is on the opposite side as the observer, negative if the object is on the same side)
Let's calculate the image position for each lens separately and then find the overall image position.
For the first lens:
f1 = 15.1 cm
u1 = 30.0 cm
Using the lens formula, we have:
1/f1 = 1/v1 - 1/u1
Plugging in the values:
1/15.1 cm = 1/v1 - 1/30.0 cm
Simplifying:
1/v1 = 1/15.1 cm + 1/30.0 cm
Finding the common denominator and adding:
1/v1 = (2 + 1) / (30.0 cm)
Simplifying further:
1/v1 = 3 / (30.0 cm)
Now, solving for v1:
v1 = (30.0 cm) / 3
v1 = 10.0 cm
Next, we need to consider the second lens:
f2 = 15.1 cm
u2 = v1
Using the lens formula again:
1/f2 = 1/v2 - 1/u2
Plugging in the values:
1/15.1 cm = 1/v2 - 1/10.0 cm
Simplifying:
1/v2 = 1/15.1 cm + 1/10.0 cm
Finding the common denominator and adding:
1/v2 = (2 + 3) / (30.0 cm)
Simplifying further:
1/v2 = 5 / (30.0 cm)
Now, solving for v2:
v2 = (30.0 cm) / 5
v2 = 6.0 cm
Since v2 is positive, the final image is formed on the same side as the observer. Therefore, the final image is formed 6.0 cm in front of the second lens.
To determine the overall magnification of the system, we can use the magnification formula:
M = -(v2 / u1)
Plugging in the values:
M = -(6.0 cm / 30.0 cm)
Simplifying:
M = -0.2
Therefore, the magnification of the system is -0.2.