Respuesta :
Answer:
A larger image is produced when [tex]d_i[/tex] > [tex]d_o[/tex]
A smaller image is produced when [tex]d_i[/tex] < [tex]d_o[/tex]
An upright image is produced when m is positive
An upright image is produced when m is negative
Explanation:
The mirror equation is given as follows;
[tex]\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}[/tex]
[tex]m =-\dfrac{d_i}{d_o} = \dfrac{h_i}{h_o}[/tex]
For concave mirrors, f = focal length
[tex]d_i[/tex] = Image distance from the mirror (-ve [tex]d_i[/tex] = Image is behind the mirror +ve [tex]d_i[/tex] = Image is in front of the mirror)
[tex]d_o[/tex] = Object distance from the mirror (-ve [tex]d_o[/tex] = Object is behind the mirror +ve [tex]d_o[/tex] = Object is in front of the mirror)
m = Magnification (-ve m = Inverted image +ve m = upright image)
[tex]h_i[/tex] = Image height
[tex]h_o[/tex] = Object height
f = Focal length of the mirror
To produce a larger image [tex]d_i[/tex] > [tex]d_o[/tex]
To produce a smaller image [tex]d_i[/tex] < [tex]d_o[/tex]
To produce an upright image, m should be positive hence, [tex]d_i[/tex] will be negative or the image will appear behind the mirror
To produce an inverted image, m should be negative hence, [tex]d_i[/tex] will be positive or the image will form in front of the mirror.
