What is the height of the image (hi) and distance of the image (di) produced by a 4.0 cm object placed 10.0 cm from a CONCAVE MIRROR with a focal...

The mirror equation is given as:

`1/f = 1/d_i + 1/d_o`

and the magnification equation is given as:

`M = -d_i/d_o = h_i/h_o`

where, f is the focal length, di and do are distances of the image and the object from the mirror, hi and ho are heights of image and the object and M is the magnification.

In this case, f = 20.0 cm, do = 10.0 cm and ho= 4.0 cm and we have to find hi and di.

Using the mirror equation:

1/20 = 1/di + 1/10

or, 1/di = 1/20 -1/10 = -1/20

or, di = -20.0 cm (negative sign indicates a virtual image, behind the mirror).

Using the magnification equation:

-(-20)/10 = hi/4 = 2

or, hi = 2 x 4 = 8.0 cm.

Hence the image will be formed at a distance of 20 cm behind the mirror (a virtual image) and will have a height of 8 cm.

Hope this helps.