Position Barlow Lens for 3x Image w/ Focal Length -0.05m

[recon9]

1. I am trying to figure out a problem involving a barlow lens in a telescope. All I know is that the focal length of the lens is -0.05m. I want to know where to place the barlow lens in front of the image created by the first lens (as if the barlow wasnt there) in order to get an image 3x the size as the image created by the first lens(as if the barlow wasnt there)



2. I know that (1/S)+(1/S')=(1/f) and m = -(S'/S).



3. I drew a diagram and filled in some variables then tried to equate the above solutions but that gave me an answer with an unknown still in it.

 

[member 553083]

4. To solve this problem, we can use the formula m = -(S'/S)+(1/f). This tells us that S' = -mS + (1/f). We can then use this equation to calculate S' given the focal length of the lens (f) and the magnification (m) of the barlow lens. In this case, f = -0.05m and m = 3. Therefore, S' = -(3)(S) + (1/-0.05) = -3S + 20. So the barlow lens should be placed 20 cm in front of the image created by the first lens to get an image 3x the size as the image created by the first lens.