A converging lens with a focal length of 30 cm is used to create an image of a 2.0 mm long ant. The ant is placed perpendicularly to the optical axis of the lens and the length of the ant is treated as the height(a) If the lens is placed so that the image of the ant is 8mm long, upright, and viewed by looking through the lens, how far away from the ant was the lens placed?
(b) If the lens is placed so that the image of the ant is 8mm long, inverted, and viewed on a screen held some unspecified distance on the other side of the lens to the ant, how far away from the ant was the lens placed?

Question

A converging lens with a focal length of 30 cm is used to create an image of a 2.0 mm long ant. The ant is placed perpendicularly to the optical axis of the lens and the length of the ant is treated as the height(a)  If the lens is placed so that the image of the ant is 8mm long, upright, and viewed by looking through the lens, how far away from the ant was the lens placed?
(b)   If the lens is placed so that the image of the ant is 8mm long, inverted, and viewed on a screen held some unspecified distance on the other side of the lens to the ant, how far away from the ant was the lens placed?

anonymous · Answer

They said to use both ray diagram and the thin lens equation to solve the problems but I'm not sure where to start

anonymous · Answer

|dw:1348068390394:dw|

anonymous · Answer

1/f = 1/do +1/di
Mag. = hi/ho =  4 =-di/do

anonymous · Answer

|dw:1348069134730:dw|

anonymous · Answer

di is going to be negative here because the image is on the same side of the lens as the object...
multiply the 1st equation by di:
di/f = di/do +1     ...you know di/do  and f  so you can solve for di

anonymous · Answer

@Algebraic! how do we solve for question b

anonymous · Answer

|dw:1348079310766:dw|

anonymous · Answer

same way pretty much, magnification is -4, di is positive this time...