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A 15.0 cm object is 12.0 cm from a concave mirror that has a focal length of 4.8 cm. Its image is 8.0 cm in front of the mirror. What is the approximate height of the image produced by the mirror? –4 cm –10 cm 10 cm 4 cm

Respuesta :

it's                      -10                  yo

the              right          answer

Explanation :

Given that,

Height of object, h = 15 cm

Object distance, u = -12 cm (negative for concave mirror )

Focal length of concave mirror, f = -4.8 cm (negative for concave mirror)

Image distance, v = -8 cm

Height of the image, h' = ?

Using the definition of magnification :

[tex]m=\dfrac{h'}{h}=\dfrac{-v}{u}[/tex]

[tex]h'=\dfrac{-vh}{u}[/tex]

[tex]h'=\dfrac{-8\times 15}{12}[/tex]

h' = -10 cm

So, the height of the image is 10 cm and it shows inverted image.

Hence, the correct option is (B) " -10 cm "

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