Respuesta :
Answer:
p = 22.5 cm
Explanation:
For this exercise we must 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.
Let's start with the far point, the object is very far away (p = ∞) and the image must be formed at the far point of view of the person q = 45.0 cm
since the image is on the same side as the object according to the sign convention the distance is negative
[tex]\frac{1}{f} = \frac{1}{\infty } + \frac{1}{-45}[/tex]
f = -45.0 cm
now let's use the near point (q = 15.0 cm) at what distance the object should be
[tex]\frac{1}{p} = \frac{1}{f} - \frac{1}{q}[/tex]
[tex]\frac{1}{p} = \frac{1}{-45} - \frac{1}{-15}[/tex]1 / p = 1 / -45 - 1 / -15
[tex]\frac{1}{p} = - \frac{1}{45} + \frac{1}{15}[/tex] 1 / p = -1/45 + 1/15
[tex]\frac{1}{p}[/tex] = 0.0444
p = 22.5 cm
this is the closest distance you can see an object clearly
