Question #93d8b
1 Answer
Explanation:
First, let me give you 2 equations which are going to be useful in helping us solve this problem
Where
The image formed is a real image and is double in size compared to the object. One thing to understand about real images is that when you have a real image, it is always inverted (upside down) .
![https://useruploads.socratic.org/4dJ2edjiTf2OjqcToXFW_microscope-diagram-and-functions-ray_diagram_sample.jpg)
If we know the image is double in size, this means its magnification is
#m = -(i)/o# #-2 = -((cancel+i))/((cancel+o))# #-2 = -(i)/o#
We only are given focal length and nothing else.
But, we can use the magnification equation to express what the
#-2 = -(i)/o# #-2o = -i# #cancel-2o = cancel-i# #2o = i#
Now, take the thins lens equation and express
Note: We don't have to convert
-
#1/12 = 1/o + 1/i# -
#1/12 = 1/o + 1/(2o)# -
#1/12 = 2/(2o) +1/(2o)# -
#1/12 = 3/(2o)# #(2o)/12 = 3# #2o = 3*12# #2o = 36# #o = (36)/(2)# #o = 18#
Answer =