The magnification of image is 1.63 and image is formed behind the mirror at a distance of 11.45 cm, with real and upright configuration.
Given data:
The focal length of diverging lens is, f = 18.0 cm = 0.18 m.
The distance of insect (object) is, u = 7.00 cm = 0.07 m.
The given problem is based on the concept of lens equation. The mathematical formula for the lens equation is given as,
[tex]\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}[/tex]
here, v is the image distance.
Solving as,
[tex]\dfrac{1}{18}=\dfrac{1}{v}+\dfrac{1}{7}\\\\\dfrac{1}{v}=\dfrac{1}{18}-\dfrac{1}{7}\\\\\dfrac{1}{v}=\dfrac{7-18}{18 \times 7}\\\\v = -11.45 \;\rm cm[/tex]
Now, the magnification of image formed is given as,
[tex]m =\dfrac{-v}{u}\\\\m =\dfrac{-(-11.45)}{7}\\\\m =1.63[/tex]
Thus, we can conclude that the magnification of image is 1.63 and image is formed behind the mirror at a distance of 11.45 cm, with real and upright configuration.
Learn more about the magnification of image here:
https://brainly.com/question/12751191
