Question 23

[paulsmerek5356]

From #23:

II. Correct. A virtual, erect, and diminished image is produced by lens 1 (since lens 1 is a diverging lens), and this image is the object for lens 2

From exam 2 question #2:

Explanation

PHY 11.3

In order for the eyepiece to magnify the first image formed by the objective, this image has to be real. If it were virtual, no light rays would emerge from it and therefore it could not be magnified by the eyepiece.

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[quadalpha]

[GraemeMatt8033]

it's passing through the lens, to the other side, how is the image virtual? I is always positive on the opposite side of the lens than the image.

[jvuofm1709]

It doesn't matters whether the image is virtual or not. The virtual image is caused by the spreading refracted light rays from the diverging lens. Since the light rays are spread out on the other side of the diverging lens, the human eye will treat the spread rays as though they were formed from an object source. So the eye will see a smaller, closer, and erect image. This is how glasses work for people who are nearsighted; they look through a diverging lens which spreads the light rays, allowing the person to "see" the virtual image. Though the image is not real, it doesn't matter, the brain will still interpret the rays as an image.

In this example, though the image produced by the diverging lens is virtual, light rays are still spreading out from the other side of the lens. These rays are then refracted by the converging lens.

[j5devries4378]

How is lens 1 diverging?
Don't diverging lenses look like: )(
Converging lenses look like: ()?