Use ray diagrams and the mirror equation to locate the position, orientation, an

Question

Use ray diagrams and the mirror equation to locate the position, orientation, and type of image of an object placed in front of a convex mirror of focal length 20 cm.

Part A

The object distance is 190 cm , find the distance from the image to the mirror (positive for real image and negative for virtual image).

Part B

The object distance is 190 cm , find the magnification (positive for upright image and negotive for inverted image).

Part C

The object distance is 44 cm , find the distance from the image to the mirror (positive for real image and negative for virtual image).

Part D

The object distance is 44 cm , find the magnification (positive for upright image and negative for inverted image

Part E

The object distance is 11 cm , find the distance from the image to the mirror (positive for real image and negative for virtual image).

Part F

The object distance is 11 cm , find the magnification (positive for upright image and negative for inverted image).

Explanation / Answer

Part A:

mirror equation, 1/f = 1/u + 1/v

1/[-20] = 1/190 +1/v

v = -18.1 cm (virtual)

the distance from the image to the mirror = -18.1 cm (virtual)

Part B:

m = -v/u = -[-18.1/190] = 0.095 Upright Image

Magnification = 0.095, upright

Part C:

1/44 + 1/v = 1/-20

v = -13.75 cm, virtual

the distance from the image to the mirror, v = -13.75, virtual

Part D:

m = -v/u = -[-13.75/44] = 0.3125, upright

Part E:

1/11 + 1/v = 1/-20

v = -7.1 cm, inverted

Part F:

m = -v/u = -[-7.1/11] = 0.65, upright

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