Question #93d8b

First, let me give you 2 equations which are going to be useful in helping us solve this problem

#color(white)(aaaaaa)# Thin Lens Equation#color(white)(aaaa)# magnification equation
#color(white)(aaaaaaaa)##1/f = 1/o + 1/i##color(white)(aaaaaaaaa)##m = -(i)/o#

Where

#f = "focal length"#
#o = "object distance from lens"#
#i = "image distance from lens"#
#m = "magnification"#

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 #2#. We also know the image is real so its image distance is positive. If the object is placed on the opposite side of the lens,opposite to the where the image is located, its given that the object distance, o, is positive. Knowing all of this, we conclude the magnification is #-2#.

• #m = -(i)/o#
• #-2 = -((cancel+i))/((cancel+o))#
• #-2 = -(i)/o#

We only are given focal length and nothing else.

#1/12 = 1/o + 1/i#

But, we can use the magnification equation to express what the #i# variable is (image distance) in terms of the object distance #(o)# since in an algebraic equation, you can only solve for the unknown expressing only 1 variable.

• #-2 = -(i)/o#
• #-2o = -i#
• #cancel-2o = cancel-i#
• #2o = i#

Now, take the thins lens equation and express #i# in terms of #o# and solve for #o#.

Note: We don't have to convert #cm# to #m# as long as all the values are in the same unit measurement.

• #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 = #o = 18 cm#