Question #3171d

Let #p# be the object distance, #q# be the image distance, #f# be the focal length of the lens, #M# is the magnification of the lens.
Then the lens equation is as follows : #1/p+1/q=1/f#
and the magnification is #M=-q/p#.

These equations must be used in conjunction with the normal sign conventions, which are as follows:
#p>0# if object in front of lens, #p<0# if object is behind lens.
#q>0# if image is real, #q<0# if image is virtual.
#M>0 #if image is upright, #M<0# if image is inverted.
#|M|>1 # if image is enlarged M times, #|M|<1# if image is diminished M times.
#f>0# if lens is convex, #f<0# if lens is concave.

So in this case we use the given information to fist of all find the focal length of the given convex lens :

#1/36+1/12=1/f#

#thereforef=9cm#

We now find the image distance for each given object distance by repeated use of the lens equation, and the nature of the object by repeated use of sign conventions and the magnification equation:

6cm object located 30cm from convex lens of focal length 9cm:

#1/30+1/q=1/9# #=>q=12,857cm#.
#therefore M=-q/p=-12.857/30=-0.42856#

Hence this image is real, inverted, diminished, 2,571cm in height and located 12.857cm behind the lens.

6cm object located 24cm from convex lens of focal length 9cm:

#1/24+1/q=1/9# #=>q=14.4cm.#
#M=-q/p=-14.4/24=-0.6#

So this image is real, inverted, diminished, 3.6cm in height, 14.4cm behind the lens.

The others may be done in a similar way and I leave the details as an exercise for the readers... ;)