A2.70-cm-high object is situated 16.1 cm in front of a concave mirror that has a radius of curvature of 12.8 cm. calculate (a) the location and (b) the height of the image.

a) The image is formed 10.62 cm behind the mirror.

b) The image will be 1.78 cm and the negative sign shows that the height will upside down.

Explanation:

for f is the focal length of the mirror, u is the distance from the mirror to the object, v is the distance from the mirror to the image and r is the radius of curvature.

the focal length is given by:

f = -r/2 = -(12.8)/2 = -6.4 cm

a) the mirror formula:

1/f = 1/v + 1/u

1/v = 1/f - 1/u

    = 1/(-6.4) - 1/(-16.1)

    = -0.094138

 v = -10.62 cm

Therefore, the image is formed 10.62 cm behind the mirror.

b) let m be the magnification of the mirror and h1 be the height of the image, then:

m =  -v/u = h1/h

-v/u = h1/h

h1 = (-v/u)(h)

   = -(-10.62/-16.1)(2.70)

   = - 1.78 cm

Therefore, the image will be 1.78 cm and the negative sign shows that the height will upside down.

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