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 answer these questions, 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,
u is the object distance.

We are given the object distance, so we can calculate the image distance and then use it to find the height of the image.

(a) To find the image distance:

1/f = 1/v - 1/u

We know that the far point of the nearsighted person is 5.8 m, which means their near point is at infinity.

Since the person's contacts enable her to see distant objects clearly, we can assume the contacts correct her vision to have a far point at infinity. This means the focal point (f) of the contacts is at infinity, and we can substitute f = ∞ into the lens formula:

1/∞ = 1/v - 1/u

Since 1/∞ = 0, the lens formula simplifies to:

0 = 1/v - 1/u
1/v = 1/u
v = u

Therefore, the image distance (v) is equal to the object distance (u), which is 18.7 m since the tree is located at that distance.

So, the image distance is 18.7 m.

(b) To find the height of the image:

We can use the magnification equation:

magnification = image height / object height

The magnification can be calculated using the ratio of the image distance (v) to the object distance (u):

magnification = -v / u

Now we can substitute the values:

magnification = -(18.7 m) / (18.7 m) = -1

Since the magnification is negative, it indicates an inverted image.

The height of the image (h') can be calculated using the magnification equation:

magnification = h' / h

Given that the object height (h) is 1.8 m, we can rearrange the formula to solve for the image height (h'):

h' = magnification * h

h' = (-1) * (1.8 m) = -1.8 m

Since the image height is negative, it indicates an inverted image. Therefore, the height of the image formed by the contacts is 1.8 m and inverted.