Question

Two systems are formed from a converging lens and a diverging lens, as shown in parts a and b of the drawing. (The point labeled "Fconverging" is the focal point of the converging lens.) An object is placed inside the focal point of lens 1 at a distance of 8.59 cm to the left of lens 1. The focal lengths of the converging and diverging lenses are 15.00 and −20.0 cm respectively. The distance between the lenses is 50.0 cm. Determine the final image distance for each system, measured with respect to lens 2.

(a) di2 =
(b) di2 =

Answer 1

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