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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 the image if the focal length of the lens is 11.0 cm? Position:
Height:
Orientation:
The object is upright or The object is inverted.

Respuesta :

Answer:

Height: 0.62 cm

Orientation: Inverted.

Explanation:

Object distance = -45 cm = u

focal length = 11 cm

[tex]\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[/tex]

[tex]\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[/tex]

Height: 0.62 cm

Orientation: Inverted.

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