Diverging and Converging A diverging lens with a focal length of -15 cm and a co
ID: 1384979 • Letter: D
Question
Diverging and Converging A diverging lens with a focal length of -15 cm and a converging lens with a focal length of 15 cm have a common central axis. Their separation is 12 cm. An object of height 1.0 cm is 10 cm in front of the diverging lens, on the common central axis.
(a) Where does the lens combination produce the final image of the object (the one produced by the second, converging lens)? (in cm)
(b) What is the height of that image? (in cm)
(c) Is the image real or virtual?
(d) Does the image have the same orientation as the object or is it inverted?
Explanation / Answer
A) f1 = -15 cm
u1 = 10 cm (object distance for first lense)
Apply,
1/u1 + 1/v1 = 1/f1
1/v1 = 1/f1 - 1/u1
1/v1 = 1/(-15) - 1/10
v1 = -6 cm (image distance for first lense)
magnification, m1 = -v1/u1
= -(-6)/10
= 0.6
object disatnce for second lense, u2 = -(12+6) = -18 cm
f2 = 15 cm
Again Apply
1/u2 + 1/v2 = 1/f2
1/v2 = 1/f2 - 1/u2
1/v2 = 1/15 - 1/(-18)
v2 = 8.18 cm ((image distance for second lense)
from the first lanse 5.1225 + 3.95 = 9.0725 m <<<<---Answer
m2 = -v2/u2
= -(8.18)/(-18)
= +0.454
Overall magnification, M = m1*m2
= 0.6*0.454
= 0.27
b)
Image height = M*object height
= 0.27*1
= 0.27 cm <<<<---Answer
c) The image is real <<<<---Answer
d) THe image have same orientation <<<<---Answer