The figure below shows an object placed a distance
do1
from one of two converging lenses separated by
s = 1.00 m. The first lens has focal length f1 = 24.0 cm, and the second lens has focal length f2 = 46.0 cm. An image is formed by light passing through both lenses at a distance
di2 = 16.0 cm to the left of the second lens. (Include the sign of the value in your answers.)
(a) What is the value of do1 that will result in this image position?
(b) Is the final image formed by the two lenses real or virtual?
(c) What is the magnification of the final image?
(d) Is the final image upright or inverted?
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www.webassign.net/katzpse1/38-p-076.png
To find the value of do1 that will result in the given image position, we can use the lens formula:
1/f = 1/do - 1/di
Where:
- f is the focal length of the lens
- do is the object distance from the lens
- di is the image distance from the lens
We can start by finding the image distance di1 from the first lens:
f1 = 24.0 cm (given focal length of the first lens)
di1 = ?
do1 = do1 (given object distance from the first lens)
s = 1.00 m (separation between lenses)
Using the lens formula for the first lens:
1/f1 = 1/do1 - 1/di1
Rearranging the formula:
1/di1 = 1/do1 - 1/f1
Now, let's substitute the given values:
1/di1 = 1/do1 - 1/24.0
Next, let's find the image distance di2 from the second lens:
f2 = 46.0 cm (given focal length of the second lens)
di2 = 16.0 cm (given image distance from the second lens)
s = 1.00 m (separation between lenses)
Using the lens formula for the second lens:
1/f2 = 1/di2 - 1/do2
Rearranging the formula:
1/do2 = 1/di2 - 1/f2
Now, let's substitute the given values:
1/do2 = 1/16.0 - 1/46.0
Since the two lenses are separated by 1.00 m, we know that the sum of the image distances from the two lenses equals the separation distance:
di1 + di2 = s
Substituting the known values:
di1 + 16.0 cm = 1.00 m
Now, we can solve for di1:
di1 = 1.00 m - 16.0 cm
di1 = 1.00 m - 0.16 m
di1 = 0.84 m
Substituting the image distance di1 into the first lens formula to find do1:
1/di1 = 1/do1 - 1/f1
1/0.84 = 1/do1 - 1/24.0
Solve for do1:
1/do1 = 1/0.84 + 1/24.0
1/do1 = (24.0 + 0.84)/(0.84 * 24.0)
1/do1 = 24.84/(0.84 * 24.0)
do1 = 0.84 * 24.0/24.84
do1 ≈ 0.818 m
Answer:
(a) The value of do1 that will result in the given image position is approximately 0.818 m.
Now let's move on to part (b).
To determine if the final image formed by the two lenses is real or virtual, we need to consider the signs of the image distances di1 and di2.
In this case, since the image distance di2 is given as "-16.0 cm to the left of the second lens," the negative sign indicates that the image formed by the second lens is virtual.
Therefore, the final image formed by the two lenses is virtual.
Answer:
(b) The final image formed by the two lenses is virtual.
Moving on to part (c).
To calculate the magnification of the final image, we can use the magnification formula:
m = -di2/do1
Substituting the given values:
m = -(-16.0 cm)/(0.818 m)
Simplifying:
m = 16.0 cm/0.818 m
m ≈ 19.57
Answer:
(c) The magnification of the final image is approximately 19.57.
Now let's move on to the final part, (d).
To determine if the final image formed by the two lenses is upright or inverted, we need to consider the magnification value.
Since the magnification, m, in this case, is positive (19.57), it indicates that the image formed is upright.
Answer:
(d) The final image formed by the two lenses is upright.
To find the value of do1 that will result in the given image position, we can use the lens formula and apply it for both lenses. The lens formula relates the object distance (do), image distance (di), and focal length (f) of a lens:
1/f = 1/do + 1/di
We know the focal lengths of both lenses (f1 = 24.0 cm and f2 = 46.0 cm) and the image distance di2 = 16.0 cm. We can substitute these values into the formula for each lens and solve for do1.
For the first lens (lens 1):
1/f1 = 1/do1 + 1/di1
Since the light passes through both lenses, the image distance di1 of the first lens becomes the object distance do2 of the second lens.
For the second lens (lens 2):
1/f2 = 1/do2 + 1/di2
Now we can substitute the given values into these equations and solve for do1.
(a) To find do1 that will result in the given image position:
1/24 = 1/do1 + 1/16
Solving this equation will give the value of do1.
(b) To determine if the final image formed by the two lenses is real or virtual, we analyze the sign of the image distance (di) of the final image. If di is positive, the image is real, and if di is negative, the image is virtual. In this case, the given image distance di2 = 16.0 cm is stated to be "to the left of the second lens," indicating that it is negative. Therefore, the final image formed by the two lenses is virtual.
(c) The magnification of the final image can be calculated using the magnification formula, which is given by:
Magnification (m) = - di/do
Substitute the given values of di2 and do1 into this formula to find the magnification of the final image.
(d) To determine if the final image formed by the two lenses is upright or inverted, we analyze the sign of the magnification. If the magnification is negative, the image is inverted, and if the magnification is positive, the image is upright. Calculate the magnification as explained in (c) and analyze its sign to determine if the final image is upright or inverted.