Question

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 24.0 m away and 2.9 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?

Answer 1

The image distance of the tree viewed through the contacts is approximately 5.97 m. The height of the image formed by the contacts is approximately 0.71843 m.

Step-by-step explanation:

The image distance of the tree viewed through the contacts can be found using the lens formula:

1/f = 1/v - 1/u

Where f is the focal length of the lens, v is the image distance, and u is the object distance. Since the person can see distant objects clearly, the focal length of the contacts is equal to the far point distance, which is 5.8 m. The object distance, in this case, is the distance of the tree, which is 24.0 m. Plugging in these values, we can solve for v:

1/5.8 = 1/v - 1/24.0

Simplifying the equation gives:

v = 5.97 m

So, the image distance of the tree viewed through the contacts is approximately 5.97 m.

To find the height of the image formed by the contacts, we can use the magnification formula:

m = -v/u

Where m is the magnification, v is the image distance, and u is the object distance. Plugging in the values, we get:

m = -5.97/24.0

Simplifying the equation gives:

m = -0.2487

Since the magnification is negative, the image formed by the contacts is inverted. To find the height of the image, we can multiply the height of the tree by the magnification:

Image height = m * object height = -0.2487 * 2.9 = -0.71843 m

So, the height of the image formed by the contacts is approximately 0.71843 m.