Answer-(a) For the system in part (a), we can use the thin lens equation to find the image distance for lens 1:
1/f = 1/do + 1/di1
where f is the focal length of the converging lens, do is the object distance (the distance from the object to the lens), and di1 is the image distance (the distance from the image to the lens).
Plugging in the given values, we get:
1/15 = 1/8.59 + 1/di1
Solving for di1, we find:
di1 = 22.37 cm
Now we can use the image from lens 1 as the object for lens 2. The distance from the image to lens 2 is:
do2 = 50.0 cm - di1
do2 = 27.63 cm
Using the thin lens equation again, we can find the final image distance:
1/f = 1/do2 + 1/di2
where f is the focal length of the diverging lens, do2 is the object distance (the distance from the object to the lens), and di2 is the image distance (the distance from the image to the lens).
Plugging in the given values, we get:
1/-20 = 1/27.63 + 1/di2
Solving for di2, we find:
di2 = -33.53 cm
(b) For the system in part (b), we can follow the same process as in part (a), but with different values.
Using the thin lens equation for lens 1, we find:
1/15 = 1/8.59 + 1/di1
Solving for di1, we get:
di1 = 22.37 cm
The distance from the image to lens 2 is:
do2 = 50.0 cm - di1
do2 = 27.63 cm
Using the thin lens equation for lens 2, we find:
1/-20 = 1/27.63 + 1/di2
Solving for di2, we get:
di2 = -13.82 cm