An object is located a distance x_0 = 7.9 cm in front of a concave mirror with a radius of curvature r = 22.2 cm. (a) write an expression for the image distance, d_i. (b) numerically, what is the image distance, d_i, in centimeters? (c) is this a real or virtual image?

a. d_i = \frac{fx_o}{x_o-f}

b. d_i = -27.4cm

c. The image is virtual.

Explanation:

(a)Using the mirror formula,

\frac{1}{d_i}+\frac{1}{x_o}=\frac{1}{f}  

\frac{1}{d_i}\ =\frac{1}{f}-\frac{1}{x_o}

\frac{1}{d_i}\ =\frac{x_o-f}{fx_o}

d_i =\frac{fx_o}{x_o-f}

(b) focal length(f)= r/2

      = 22.2cm/2 = 11.1cm

using real is positive sign convention, the focal length of a concave mirror is positive. hence its 11.1cm

d_i=\frac{11.1* 7.9}{7.9-11.1}

d_i=\frac{87.68}{-3.2}

d_i= -27.4cm

(c)from real is positive sign convention the image is a virtual image since our image distance is negative (-27.4cm).

the answer is a. at the equator