What distance would an object need to be located in front of a converging lens for the image to be the same size as the object?
Closer than F
Between F and 2F
Beyond 2F
At 2F

2*F

Explanation:

If we put an object of a given size exactly at a distance 2*F from the lens, the virtual image (the image generated by the lens) will be generated at a distance 2*F from the lens and the size will be equal to the size of the real object (but the image will be inverted)

Now let's do the math.

The relation between the distance of the object to the lens O, and the distance between the image and the lens I is:

1/O + 1/I = 1/F

solving for O, we get:

1/O = 1/F - 1/I = (I - F)/(F*I)

O = F*I/(I - F)

Such that the relation between the height of the original object, H and the height of the virtual image H' is:

H/H' =  -I/O

Replacing by O we get:

H/H' = -I/(F*I/(I - F))

If the sizes are equal, then H/H' = - 1  (remember that the image is inverted, thus the sign)

-1 = -I/(F*I/(I - F))

F*I/(I - F) = I

F*I = (I - F)*I

F = (I - F)

F + F = I = 2*F

The distance between the image and the lens is 2*F

O =  F*I/(I - F) = F*2*F/(2*F - F) =  2*F

The object is at a distance 2*F from the lens.

they are playing too much beyblade in the office

explanation:

one "light year" is the distance that light travels, through vacuum, in one julian year.   that's 365.25 days, of 24 hours each.