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Harlamova29_29 [7]
4 years ago
15

An object is 45.0 cm from a converging lens and the object is 1.90 cm tall. What is the position and height and orientation of t

he image if the focal length of the lens is 11.0 cm? Position:
Height:
Orientation:
The object is upright or The object is inverted.
Physics
1 answer:
Digiron [165]4 years ago
3 0

Answer:

Height: 0.62 cm

Orientation: Inverted.

Explanation:

Object distance = -45 cm = u

focal length = 11 cm

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}\\\Rightarrow \frac{1}{v}=\frac{1}{11}+\frac{1}{-45}\\\Rightarrow v=\frac{11\times 45}{45-11}\\\Rightarrow v=14.6\ cm

\frac{h_i}{h_o}=\frac{d_i}{d_o}\\\Rightarrow h_i=h_o\times \frac{d_i}{d_o}\\\Rightarrow h_i=1.9\times \frac{14.6}{-45}\\\Rightarrow h_i=-0.62 cm

Height: 0.62 cm

Orientation: Inverted.

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Answer:

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Explanation:

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Working a little the equation, we can take:

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(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

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