Question

Al colocar un objeto delante de una lente convergente se genera una imagen real invertida y de doble tamaño, sabiendo que dicha imagen se forma a 30cm delante de la lente, calcule la distancia focal.
Aumento Lente Convergente

Answer 1

Answer:

30 cm

Explanation:

To solve this problem, we use the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we know that

q = -30 cm is the distance of the image from the lens; it is negative because the image is formed in front of the lens, so it is a virtual image

We also know that the size of the image is twice that of the object, so the magnification is 2:

$M=-\frac{q}{p}=2$

where M is the magnification of the lens. Solving this equation for p,

$p=-\frac{q}{2}=-\frac{-30}{2}=15 cm$

So, the distance of the object from the lens is 15 cm.

Now we can finally solve the lens equation to find f, the focal length:

$\frac{1}{f}=\frac{1}{15}+\frac{1}{-30}=\frac{1}{30}\\\rightarrow f=30 cm$