two lenses,one converging with focal length 20.0 cm and one diverging with focal length -10.0 cm , are placed 25.0 cm apart. An object is placed 60.0 cm in front of the converging lens. Determine
A) The position and B) the magnification of the final image formed.C)Sketch a ray diagram for the system.
A) The position of the final image is 40 cm from the converging lens.
B) The magnification of the final image is -2.
C) See the attached diagram.
To determine the position and magnification of the final image formed by the two lenses, we can use the lens formula and the lens-maker's equation.
A) Position of the final image:
1. Start by calculating the position of the image formed by the converging lens using the lens formula:
(1/f) = (1/v) - (1/u)
Where:
f = focal length of the converging lens = 20.0 cm
v = distance of the image from the converging lens (unknown)
u = distance of the object from the converging lens = -60.0 cm (negative because the object is located in front of the lens)
Rearranging the formula, we have:
(1/v) = (1/f) + (1/u)
Calculating:
(1/v) = (1/20.0) + (1/-60.0)
(1/v) = 0.05 - 0.0167
(1/v) = 0.0333
Therefore, v = 1/(0.0333) = 30.03 cm
The image formed by the converging lens is located 30.03 cm in front of the lens.
2. Next, calculate the position of the image formed by the diverging lens. Since the lenses are 25.0 cm apart, the distance from the converging lens to the diverging lens is 30.03 cm - 25.0 cm = 5.03 cm.
Using the lens formula again:
(1/f) = (1/v) - (1/u)
Where:
f = focal length of the diverging lens = -10.0 cm
v = distance of the image from the diverging lens (unknown)
u = distance of the object from the diverging lens = -5.03 cm (negative because the object is located in front of the lens)
Rearranging the formula, we have:
(1/v) = (1/f) + (1/u)
Calculating:
(1/v) = (1/-10.0) + (1/-5.03)
(1/v) = -0.1 - 0.1988
(1/v) = -0.2988
Therefore, v = 1/(-0.2988) ≈ -3.34 cm
The image formed by the diverging lens is located approximately 3.34 cm behind the lens.
Hence, the position of the final image formed is 3.34 cm behind the diverging lens.
B) Magnification of the final image:
To find the magnification of the final image, we can use the formula:
magnification (m) = v/u
Where:
v = distance of the final image from the diverging lens = -3.34 cm
u = distance of the object from the converging lens = -60.0 cm
Calculating:
magnification (m) = (-3.34)/(-60.0) = 0.0557
Therefore, the magnification of the final image formed is approximately 0.0557.
C) Ray Diagram:
To sketch a ray diagram for the system, follow these steps:
1. Draw a vertical line to represent the optical axis.
2. Place the converging lens 60.0 cm to the left of the optical axis.
3. Draw the object (an arrow, for example) 60.0 cm to the left of the converging lens along the optical axis.
4. Draw a ray from the object traveling parallel to the optical axis. This ray will pass through the focal point of the converging lens (20.0 cm to the right of the lens).
5. Draw a ray from the object traveling through the focal point of the converging lens. This ray will become parallel to the optical axis after passing through the lens.
6. The ray diagram will show the image formed by the converging lens at a distance of 30.03 cm in front of the lens.
7. Add the diverging lens 25.0 cm to the right of the converging lens.
8. Draw a ray from the image formed by the converging lens. This ray will appear to come from the position of the image.
9. Draw another ray from the image formed by the converging lens parallel to the optical axis. After passing through the diverging lens, it will appear to come from the focal point of the diverging lens.
10. The rays will diverge and won't actually meet. However, by extending the rays backward, you can determine the location of the virtual image formed by the system, which is approximately 3.34 cm behind the diverging lens.
By following these steps, you should be able to sketch a ray diagram for the system and visualize the optical path of the rays.
To determine the position and magnification of the final image formed by the system of two lenses, we can use the lens formula and magnification formula.
A) Position of the final image:
The distance between the object and the converging lens is given as 60.0 cm. Using the lens formula for the converging lens, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance, we can find the image distance for the converging lens.
1/20 = 1/v - 1/60
(3/60) - (1/60) = 1/v
2/60 = 1/v
v = 60/2
v = 30 cm
The image distance for the converging lens is 30 cm.
Next, we need to find the object distance for the diverging lens. The image formed by the converging lens acts as the object for the diverging lens. The object distance for the diverging lens is the same as the image distance from the converging lens.
u' = v = 30 cm
Now, using the lens formula for the diverging lens, 1/f' = 1/v' - 1/u', where f' is the focal length, v' is the image distance, and u' is the object distance, we can calculate the image distance for the diverging lens.
1/-10 = 1/v' - 1/30
(-3/30) - (1/30) = 1/v'
-4/30 = 1/v'
v' = -30/4
v' = -7.5 cm
The image distance for the diverging lens is -7.5 cm.
To find the position of the final image formed by the system, we consider the distance between the two lenses. Since the image distance for the diverging lens is negative, we have to account for the difference in sign. Adding the two distances:
30 cm + (-7.5 cm) = 22.5 cm
Hence, the position of the final image formed by the system of two lenses is 22.5 cm to the right of the second lens.
B) Magnification of the final image:
The magnification for each lens can be calculated using the formula magnification (m) = -v/u, where v is the image distance and u is the object distance.
For the converging lens:
m1 = -30/60
m1 = -0.5
For the diverging lens:
m2 = -(-7.5)/30
m2 = 0.25
To find the total magnification, multiply the individual magnifications:
m = m1 * m2
m = -0.5 * 0.25
m = -0.125
Therefore, the magnification of the final image formed by the system of two lenses is -0.125.
C) Ray diagram:
Unfortunately, I am unable to provide a visual diagram as I am a text-based AI. However, I can describe the ray diagram for the system.
1. Draw the converging lens, placed 60.0 cm away from the object. Label the lens as "Lens 1" and mark its focal length as 20.0 cm.
2. Draw one ray from the top of the object parallel to the principal axis of the lens. This ray will refract through the lens and pass through the focal point on the other side.
3. Draw another ray from the top of the object that passes through the center of the lens. This ray will continue undeviated.
4. Sketch the refracted rays after passing through Lens 1.
5. From the point where the refracted rays meet, draw the diverging lens with a focal length of -10.0 cm. Label it as "Lens 2".
6. Extend the rays that pass through Lens 1 until they reach Lens 2. Note that the parallel ray will appear to diverge and the one passing through the center will continue undeviated.
7. Draw the refracted rays after passing through Lens 2.
8. Locate the final image formed by the system where the rays intersect after passing through both lenses. This will be the position determined earlier (22.5 cm to the right of Lens 2).
Remember, this description should give you a general idea of the ray diagram, but it's important to refer to a diagram or seek additional resources for a more accurate visualization.