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A student holds a diverging lens 25.0 cm from the lettering on a poster on the wall. A virtual image of the letters is formed 10.5 cm between the lens and the wall. What are the focal length of the lens and orientation of the image? A) 7.4 cm and upright B) 7.4 cm and inverted C) 18.1 em and upright D) 18.1 em and inverted E) -7.4 cm and upright

Respuesta :

Answer:

-18.1 cm and upright

Explanation:

u = Object distance =  25 cm

v = Image distance = -10.5 cm

f = Focal length

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{25}+\frac{1}{-10.5}\\\Rightarrow \frac{1}{f}=\frac{-29}{525}\\\Rightarrow f=\frac{-525}{29}=-18.1\ cm[/tex]

Focal length of the lens is -18.1 cm

Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-10.5}{25}\\\Rightarrow m=0.45[/tex]

Since magnification is positive the image upright

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