An object 2.00 cm high is placed 39.0 cm to the left of a converging lens having
ID: 1520949 • Letter: A
Question
An object 2.00 cm high is placed 39.0 cm to the left of a converging lens having a focal length of 29.0 cm. A diverging lens with a focal length of -20.0 cm is placed 110 cm to the right of the converging lens. (Use the correct sign conventions for the following answers.)
(a) Determine the final position and magnitude of the final image.
image position
?
magnitude
?
(b) Is the image upright or inverted?
(c) Repeat parts (a) and (b) for the case where the second lens is a converging lens with a focal length of +20.0 cm.
image position ?cm
magnification
?
Explanation / Answer
For converging lens :
f = 29 cm
do = object distance = 39 cm
di = image distance
using the lens equation
1/f = 1/do + 1/di
1/29 = 1/39 + 1/di
di = 113.10 cm
ho = object height = 2 cm
hi = image height = (-di/do) ho = (- 113.10/39) (2) = - 5.8 cm
For the diverging lens :
f = - 20 cm
do = object distance = 113.10 - 110 = 3.10 cm
di = image distance
using the lens equation :
using the lens equation
1/f = 1/do + 1/di
1/(-20) = 1/3.10 + 1/di
di = - 2.7 cm
final image position = 110 + 2.7 = 112.7 cm to the right of converging lens
b)
the image in inverted
c)
For the second converging lens :
f = 20 cm
do = object distance = 113.10 - 110 = 3.10 cm
di = image distance
using the lens equation :
using the lens equation
1/f = 1/do + 1/di
1/(20) = 1/3.10 + 1/di
di = - 3.67 cm
final image position = 110 + 3.67 = 113.7 cm to the right of converging lens
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