An object is placed 20.0 cm from a thin converging lens along the axis of the lens. If a real image forms behind the lens at a distance of 8.00 cm from the lens, what is the focal length of the lens?

5.71cm (a.)

Well thanks noah i guess you were too late for lauryn but not for me :)

To find the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u,

where f is the focal length, v is the image distance, and u is the object distance.

Given:
u = -20.0 cm (negative sign indicates that the object is located on the opposite side of the lens from the incident light)
v = -8.00 cm (negative sign indicates that the image is formed on the same side of the lens as the incident light)

Substituting these values into the lens formula:

1/f = 1/-8.00 - 1/-20.0

Simplifying the equation:

1/f = (-1/8.00) - (-1/20.0) = -5/40.0 + 2/40.0 = -3/40.0

To isolate f, take the reciprocal of both sides:

f = -40.0/3

Therefore, the focal length of the lens is approximately -13.3 cm.

To find the focal length of a lens, we can use the lens formula, which is given by:

1/f = 1/v - 1/u,

where:
- f is the focal length of the lens,
- v is the image distance from the lens (negative if the image is formed on the same side as the object),
- u is the object distance from the lens (negative if the object is on the same side as the image).

In this case, the object is placed 20.0 cm from the lens, so u = -20.0 cm. The image forms behind the lens at a distance of 8.00 cm, so v = -8.00 cm.

Now let's substitute these values into the lens formula:

1/f = 1/-8.00 - 1/-20.0.

To solve this equation, we need to find the least common multiple of -8.00 and -20.0, which is 40.00. Simplifying the equation further:

1/f = -5/40.00 + 2/40.00,
1/f = -3/40.00.

Now, invert both sides of the equation to solve for f:

f = -40.00/3.

Therefore, the focal length of the lens is -13.3 cm.