A 2.11 cm high insect is 1.31 m from a -125 mm focal-length lens. Where is the image?

Distance of the image is ?
How high is it?
What type is it?

Find the image distance from

1/do + 1/di = 1/f,

1/1.31 + 1/di = 1/(-125•10^-3).
di = - … (m) – virtual, in front of the length
The height of the image may be found from

M =h1/h2 = - di/do,
h1 = - … (inverted)

Thank you Elena, I had virtual and inverted also, but the book says it is virtual and upright. so I thought everything else was worry. Maybe it is the book?

Anybody's guess...

To find the image location, we can use the lens formula:

1/f = 1/v - 1/u

where:
- f is the focal length of the lens,
- v is the distance of the image from the lens, and
- u is the distance of the object from the lens.

Using the given values, we can plug them into the formula:

1/-0.125 = 1/v - 1/1.31

To simplify this equation, we need to take the reciprocal of both sides:

-8 = v - 1/1.31

Next, we solve for v:

-7.69 = v

The distance of the image from the lens is approximately -7.69 meters. Note that the negative sign indicates that the image is formed on the same side as the object (which is opposite to the conventional side of a lens).

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

magnification = -v/u

where the magnification is the ratio of the image height to the object height.

Given that the object height is 2.11 cm (0.0211 m) and we have already found the value of v, we can calculate the magnification:

magnification = -(-7.69)/1.31 ≈ 5.87

This means that the height of the image is 5.87 times the height of the object. Therefore, the height of the image is:

0.0211 m * 5.87 ≈ 0.1239 m

So, the height of the image is approximately 0.1239 meters.

To determine the type of image formed, we can use the magnification value. Since the magnification is positive, we know that the image is upright. Additionally, since the image is formed on the same side as the object (as indicated by the negative sign in the image distance), the image is virtual.

Therefore, the image is upright and virtual.