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3 years ago

9

Frente a una lente convergente delgada se coloca un objeto a una distancia de 50 cm. La imagen de este objeto aparece del otro l

ado a 80 cm de la lente. ¿Cuál es la distancia focal de la lente?

1 answer:

katovenus [111]3 years ago

4 0

Answer:

La distancia focal del objetivo es de aproximadamente 30,77 cm

Explanation:

Los parámetros dados son;

La distancia del objeto desde la lente, u = 50 cm

La distancia de la imagen desde la lente, v = 80 cm

La fórmula del espejo se puede escribir de la siguiente manera;

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

Dónde;

f = La distancia focal de la lente

v = La distancia de la imagen desde la lente

u = La distancia del objeto desde la lente

Sustituyendo los valores da;

$\dfrac{1}{f} = \dfrac{1}{80} + \dfrac{1}{50} = \dfrac{80 + 50}{4000} = \dfrac{13}{400}$

$\therefore f = \dfrac{400}{13} = 30\dfrac{10}{13} \ cm$

La distancia focal de la lente, f = 30¹⁰/₁₃ cm ≈ 30,77 cm.

La distancia focal del objetivo ≈ 30,77 cm.

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