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Anastasy [175]
3 years ago
5

Un espejo convexo y un cóncavo tienen la misma distancia radial igual a 5m. La separación entre espejos es de 3 veces la distanc

ia focal. Si un objeto de altura 1.75m que se encuentra entre estos dos espejos, se refleja en ambos, determine la relación de alturas entre las imágenes que se forman.
Physics
1 answer:
NARA [144]3 years ago
8 0

Answer:

La altura de la imagen producida por el espejo cóncavo es 5 veces la altura de la imagen producida por el espejo convexo

Explanation:

La ecuación de espejo para espejos cóncavos viene dada por la relación;

\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}

Para espejos convexos, tenemos;

-\dfrac{1}{f} = \dfrac{1}{d_o} - \dfrac{1}{d_i}

Dado que la separación de los dos espejos = 3 × f

La ubicación del objeto = Entre los espejos = f × 3/2

Altura del objeto = 1.75 m

Radio de curvatura de los espejos = 5 m.

Por lo tanto, f = 5/2 = 2.5 m.

Distancia del objeto = f × 3/2 = 3.75 m.

Para el espejo cóncavo, tenemos que la distancia de la imagen se puede encontrar de la siguiente manera;

\dfrac{1}{2.5} = \dfrac{1}{3.75} + \dfrac{1}{d_i}

Lo que da;

\dfrac{2}{5} = \dfrac{4}{15} + \dfrac{1}{d_i}

\dfrac{1}{d_i} = \dfrac{2}{5} - \dfrac{4}{15} = \dfrac{2}{15}

d_i = 15/2 = 7.5 m

Que es una imagen real

m = \dfrac{-d_i}{d_o}  = \dfrac{-7.5}{3.75} = -2

Por lo tanto, la imagen es real, invertida y es dos veces la altura del objeto.

La altura de la imagen, h_i = m × Altura del objeto = -2 = -3.5 m

Para el espejo convexo, tenemos que la distancia de la imagen se puede encontrar de la siguiente manera;

-\dfrac{1}{2.5} = \dfrac{1}{3.75} - \dfrac{1}{d_i}

Lo que da;

-\dfrac{2}{5} = \dfrac{4}{15} - \dfrac{1}{d_i}

-\dfrac{1}{d_i} = -\dfrac{2}{5} - \dfrac{4}{15} = -\dfrac{2}{3}

d_i = 3/2 = 1.5 m

Que es una imagen real

m = \dfrac{-d_i}{d_o}  = \dfrac{-1.5}{3.75} = -\dfrac{2}{5}

Por lo tanto, la imagen es real, invertida y es 2/5 veces la altura del objeto

La altura de la imagen, h_i = m × Altura del objeto= -2/5 × 1.75 = -0.7 m

Por lo tanto, cuando la altura de la imagen formada por el espejo cóncavo se divide por la altura de la imagen formada por el espejo convexo, tenemos;

\dfrac{h_{(i, concave)}}{h_{(i, convex)}} =\dfrac{-3.5}{-0.7} = 5

Por lo tanto, la altura de la imagen producida por el espejo cóncavo = 5 veces la altura de la imagen producida por el espejo convexo

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A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate
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Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\

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