Question

A virtual image of a marble is located 13 cm in front of a diverging lens with a focal length of 29 cm.

Where is the marble located?
PLS ANSWER AM IN DESPERATE NEED OF HELP

Answer 1

The marble is located 16 cm behind the lens.

Step-by-step explanation:

When an object is located in front of a diverging lens (a lens with a negative focal length), a virtual image is formed. In this case, the virtual image of the marble is located 13 cm in front of the lens. The focal length of the lens is 29 cm. The marble is located at a distance of 16 cm behind the lens (13 cm in front of the lens + (-29 cm) focal length of the lens).

Answer 2

Image distance = 23.6cm

Step-by-step explanation:

Given

`Focal\ Length\ (f) = 29cm`

`Object\ Distance (u) = 13\ cm`

Required

Determine the image distance (v)

This can be calculated using:

`(1)/(f) = (1)/(u) + (1)/(v)`

`(1)/(29) = (1)/(13) + (1)/(v)`

Collect Like Terms

`(1)/(v) = (1)/(29) - (1)/(13)`

`(1)/(v) = (13 - 29)/(29 * 13)`

`(1)/(v) = (-16)/(377)`

`v = (-377)/(16)`

`v = -23.5625`

`v = -23.6\ cm`

This implies that the image distance is 23.6cm and it is a virtual image.