A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

Part A

Calculate the image position

s'=

Part B

Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.

h =

The image of an object closer to the converging lens than the focal length

is larger than the object.

Part A: The image position away from the lens and on the same side of the

lens is

Part B: The image height is

Reasons:

The given parameters are;

Height of the object = 1.0 cm-tall

Distance of the object in front of the lens, u = 13 cm

Focal length of the converging lens, f = 40 cm

Part A:

The lens formula is;

Therefore;

Which gives;

The image position, v ≈ cm. away from and on the same side as the

object relative to the lens.

Part B

The image height is given by the magnification relation;

Therefore;

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A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

And the positive sign means the image is upright.

be subject to a force andchange its motion