Answer:
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).
(a) The expression for image distance is given by [tex]d_i = \frac{fx_o}{x_o-f}[/tex]
(b) The image distance is -27.4cm
(c) The image is virtual
(a) The mirror formula is given by:
[tex]\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}[/tex]
(b) focal length of the mirror is
(f)= r/2= 22.2cm/2 = 11.1cm
The image formation is:
[tex]d_i=\frac{11.1* 7.9}{7.9-11.1}\\\\d_i=\frac{87.68}{-3.2}\\\\d_i= -27.4cm[/tex]
27.4 cm on the opposite of the mirror
(c) the image is a virtual image since the image distance is negative (-27.4cm).
Learn more about concave mirror:
https://brainly.com/question/1261867?referrer=searchResults
