You have a lens whose focal length is 12.9 cm. you place an object on the axis of the lens at a distance of 6.59 cm from it. how far is the object's image from the lens (as a positive quantity)? number cm if it can be determined, is the image real or virtual? ocannot be determined real o virtual if it can be determined, is the image upright or inverted with respect to the object? o cannot be determined o upright o inverted

(I). The distance of the image from the lens is 13.47 cm.

(I). The image is virtual.

(III). The image is upright.

Explanation:

Given that,

Focal length = 12.9 cm

Object distance = 6.59 cm

(I). We need to calculate the image distance

Using formula of lens

Where, f = focal length

v = image distance

u = object distance

Put the value into the formula

The distance of the image from the lens is 13.47 cm.

(II). Negative sign shows the image is formed in left side of the lens.

So, the image is virtual.

(III). We need to calculate the magnification

Using formula of magnification

Put the value into the formula

Positive sign show the image is upright.

Hence, (I). The distance of the image from the lens is 13.47 cm.

(II). The image is virtual.

(III). The image is upright.

this is a good question! i'll do some research and come back when i have the answer!

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down