Answer:
p = 6.64 cm
Explanation:
For this exercise we use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image
[tex]\frac{1}{p} = \frac{1}{f} - \frac{1}{q}[/tex]
1 / p = 1 / 2.2 - 1/3.29
1 / p = 0.15059
p = 6.64 cm
therefore the farthest distance from the object is 6.64 c
