Question about height of image and magnification

AI Thread Summary
The discussion centers on calculating the image height and magnification for a concave mirror with a 4.6 cm tall object located 16 cm from the center of curvature. The calculated image position is 9.6 cm, and the magnification is -0.6, leading to an image height of -2.76 cm. The negative height indicates that the image is inverted, while the negative distance suggests the image is real. Clarifications are provided that positive image distance indicates a real image, while negative distance indicates a virtual image. The conclusion emphasizes that the negative magnification confirms the image is inverted.
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Homework Statement



so if the original object is 4.6 cm tall located at x = 0 and it's a concave mirror

distance from C to the object is 16 cm and Radius of the mirror is 12 cm



The Attempt at a Solution



to find the position of x image

i got 9.6 cm

then for magnification i got -.6 which is also equal to M = -di/do = Hi /Ho

so since the object is 4.6 tall the Hi is -2.76cm

so it's inverted? and it's real? since it's inside the mirror and the object

how can Hi be negative what does this mean if the di is negative it's virtural i think so the image is real

but should the height be in magnitude but what i got from the equation was -2.76cm

please explain

thank you
 
Physics news on Phys.org
Positive di means the image is real.
Negative di means the image is virtual.

Negative magnification means the image is inverted.
Positive magnification means the image is erect.
 
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