Two converging lenses are separated by 50 cm, and have focal lengths of
f1 = 20 cm and f2 = 10 cm, as illustrated in Figure 2. An object is placed on the
left of Lens 1. The final image formed by the two lenses is located at 26.0 cm to
the right of Lens 1. (a) Determine the object position, p, measured with respect
to Lens 1 and (b) describe the final image including its magnification and
orientation.
To determine the object position, p, measured with respect to Lens 1, 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, and u is the object distance from the lens.
Given that the final image formed by the two lenses is located at 26.0 cm to the right of Lens 1, we can label this distance as v2. Since Lens 2 has a focal length of f2 = 10 cm, we can write:
1/f2 = 1/v2 - 1/u2
We also know that the distance between the two lenses is 50 cm. Therefore, the object distance from Lens 2 is u2 + 50 cm.
Now, let's calculate the image distance from Lens 2 using the lens formula:
1/f2 = 1/v2 - 1/u2
1/10 = 1/26 - 1/(u2 + 50)
Now, we need to find the object distance from Lens 1, which is the sum of the image distance from Lens 2 (v2) and the distance between the two lenses (50 cm):
u1 = v2 + 50
We have two equations with two unknowns (v2 and u2), we can solve these equations simultaneously to find the values. However, in this case, let's assume the rays are close to paraxial, which gives us:
v2 = -26.0 cm
u2 << f2
Now, substitute these values into the equation:
1/10 = 1/(-26.0) - 1/(u2 + 50)
Simplifying and solving for u2:
1/10 = -1/26 - 1/(u2 + 50)
1/10 + 1/26 = -1/(u2 + 50)
(26 + 10)/(10 * 26) = -1/(u2 + 50)
36/260 = -1/(u2 + 50)
(u2 + 50)/36 = -260/36
(u2 + 50)/36 = -13/2
Cross-multiplying:
2(u2 + 50) = -13 * 36
2u2 + 100 = -468
2u2 = -568
u2 = -284 cm
Now, substitute u2 into the equation for u1:
u1 = v2 + 50
u1 = -26.0 + 50
u1 = 24.0 cm
Therefore, the object position, p, measured with respect to Lens 1 is 24.0 cm.
Now let's move to part (b):
To describe the final image formed by the two lenses including its magnification and orientation, we need to determine the overall magnification of the system.
The magnification of a single lens is given by:
magnification = -v/u
where v is the image distance from the lens, and u is the object distance from the lens. The negative sign indicates the orientation of the image, where a negative magnification indicates an inverted image.
For Lens 1, using the object distance (p = 24.0 cm) we found previously:
magnification1 = -v1/u1
For Lens 2, using the object distance (u2 = -284 cm) and the image distance (v2 = -26.0 cm) we found previously:
magnification2 = -v2/u2
The overall magnification of the system will be the product of the magnifications of the individual lenses:
overall magnification = magnification1 * magnification2
Now, substitute the values we found earlier:
magnification1 = -v1/u1 = -(-26.0 cm)/(24.0 cm) = 1.083
magnification2 = -v2/u2 = -(-26.0 cm)/(-284 cm) = 0.092
Therefore,
overall magnification = 1.083 * 0.092 = 0.0995
The final image formed by the two lenses has a magnification of approximately 0.0995, and it is inverted since the magnification is negative.