An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. its image is 16.0 cm in front of the mirror. the height of the image produced by the mirror, to the nearest tenth, is cm.

3.2 cm

Explanation:

The magnification of a concave mirror is given by:

where, h' is the height of the image, h is the height of the object, v is the image distance and u is the object distance.

Given:

h = 8.0 cm

u= - 40.0 cm

v= - 16.0 cm

Height of the image = 3.2 cm

Diminished, real and inverted image would be formed.

h' = -3.2 cm                                

Explanation:

It is given that,

Size of the object, h = 8 cm

Object distance, u = -40 cm

Focal length of the mirror, f = -10 cm

Image distance, v = -16 cm

Let h' is the height of the image produced by the mirror. Using the formula of magnification of mirror as :

h' = -3.2 cm

So, the height of the image produced by the mirror is 3.2 cm. Hence, this is the required solution.

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