An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12 cm from it, there is a converging lens of the same focal length. A. Find the location of the final image, in centimeters beyond the converging lens.
B. What is the magnification of the final image?

A)  q = -8.488 cm ,  B)  m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

we calculate

          = - 0.1178

          q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

we substitute

          m =

          m = 0.29

the positive sign indicates that the image is right

i guess that visual aids present persuasive examples to what you are presenting, some factors would be theme (does it fit with the color, information, what you are presenting), size (can you see it, is it a big enough span of data/information), and overall look of the visual aid.

hope this : )

your answer is c! hope this : )