The far point of a nearsighted person is 5.8 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 18.7 m away and 1.8 m high.
(a) When she looks through the contacts at the tree, what is its image distance?
(b) How high is the image formed by the contacts?
To find the answers to these questions, we can use the lens formula:
1/f = 1/v - 1/u
where:
- f is the focal length of the contact lens,
- v is the image distance,
- u is the object distance.
(a) To find the image distance (v) when the tree is viewed through the contacts, we need to find the focal length (f) of the contact lens.
We are given that the far point of the nearsighted person is 5.8 m. The far point is the maximum distance at which a nearsighted person can see objects clearly. In other words, when the person focuses at infinity (distant objects), the image is formed at their far point.
Using the lens formula, we have:
1/f = 1/v - 1/u
Since the person can see the tree clearly (distant object), the object distance (u) is the distance to the tree, which is 18.7 m.
Let's plug in the known values into the lens formula:
1/f = 1/v - 1/u
Assuming the image is formed on the retina (which is the usual case when using contact lenses), the image distance (v) would be equal to the person's far point, 5.8 m.
Therefore, the equation becomes:
1/f = 1/5.8 - 1/18.7
Now, let's calculate the focal length (f) by rearranging the equation:
1/f = (18.7 - 5.8)/18.7*5.8
1/f = 12.9/18.7*5.8
1/f = 12.9/108.46
1/f ≈ 0.119
f ≈ 1/0.119
f ≈ 8.40 m
So, the focal length of the contact lens is approximately 8.40 m.
(b) To determine the height of the image formed by the contacts, we can use the magnification formula:
magnification (m) = -v/u
where:
- m is the magnification of the lens,
- v is the image distance,
- u is the object distance.
From part (a), we have found that the image distance (v) is equal to the far point of the nearsighted person, which is 5.8 m.
Again, the object distance (u) is the distance to the tree, which is 18.7 m.
Let's plug in the values into the magnification formula:
m = -v/u
m = -5.8/18.7
m ≈ -0.310
Since the magnification (m) is negative, it indicates an inverted image.
The height of the image (h') is related to the height of the object (h) by the formula:
h' = m * h
We are given that the height of the tree (h) is 1.8 m.
So, let's calculate the height of the image (h'):
h' = -0.310 * 1.8
h' ≈ -0.558 m
The negative sign indicates that the image is inverted. Thus, the height of the image formed by the contacts is approximately 0.558 m.