1. Double lens systems. a. Two converging lenses with focal lengths of 20 cm and
ID: 1392556 • Letter: 1
Question
1. Double lens systems.
a. Two converging lenses with focal lengths of 20 cm and 20 cm are placed 10 cm
apart. A 2 cm tall object is located 15 cm from the first lens as shown in the figure.
Where is the final image created by the system? What is its height and orientation
compared to the object?
b. An object is placed 15 cm from a converging lens with a focal length of 5 cm. A
second lens is placed 5 cm from the first lens. The system forms a virtual image 10
cm away from the second lens. Is the second lens converging or diverging? What is
its focal length?
c. You have a converging and a diverging lens. The diverging lens has a focal length fd
and the converging lens has a focal length fc.The incoming rays are parallel and pass first through the converging lens. Your goal is to space the lenses so that the rays emerging from the diverging lens are parallel.
i. How far apart should the lenses be?
ii. Can you have the same outcome if light first enters the diverging lens?
Explanation / Answer
Solution:
a) Focal length of 1st converging lens= F1= 20 cm
Focal length of 2nd converging lens = F2 = 20 cm
Distance between the two lenses = d = 15 cm
Height of the object = ho = 2 cm
When the object is within the focal length , the image formed by the 1st lens is Virtual upright and on the same side as the object.
1/f = 1/p + 1/q
1/q =1/f -1/p
= 1/20 - 1 /15
=>q= Image distance from the 1st lens = -60 cm
Magnification of the 1st image = m1 = -q/p = - (-60)/15 = -4
So the image distance from the 2nd lens will be 10 +60 = 70 cm
object distance for 2nd lens = p'= 70cm
focal length = f' = 20cm
1/q'= 1/f'-1/p'
= 1/20 - 1/70 =0.0357
=> q'= image distance from the 2nd lens = 28 cm
The magnification of the 2nd image = m2 = - q'/p' = -28/70
= -0.4
Final magnification = m1 * m2 = (4)(-0.4) = -1.6
=> final image is 1.6 times the original object. It is inverted as the final image is real.
Height of the final image = 1.6 * 2 =3.2 cm
The final image is formed at a distance of 28 cm on the right side of the 2nd lens . It is real image , inverted and diminished.
b) Focal length of 1st lens = f1 = 5 cm
object distance from 1st lens = p = 15 cm
distance between the two lenses = d = 5 cm
Image distance due to 1st lens = q
1/q=1/f-1/p = 1/5 -1/15 = 0.1333
=> q= distance from 1st lens to the original object = 7.5 cm
This becomes the object to the 2nd lens
object distance to the 2nd lens = 7.5- 5= 2.5 cm =p'
final image distance from the 2nd lens = q' = 10 cm
since the object is within the focal length of the 2nd lens, it forms an image which is virtual, upright and enelarged and on the same side as its object.
1/ f 2 = 1/p'+1/q'
= 1/ 2.5 - 1/10
= 0.3
=> f2 = 1/0.3 = 3.33 cm
It is a converging lens since the focal length is positive.
c) i) when the incoming rays are parallel thru the 1st lens (converging lens ) the object is situated at Infinity and the image is formed at the focal point ; So the image distance from the 1st lens = fc.
Let distance between the two lenses be L
Then the object distance for the diverging lens = L - fc
Since the final emergent rays should be parallel to the diverging lens, the final image is formed at the focal point of the diverging lens and on the same side as its object.So we should arrange the distance between the lenses in such a way that the object distance for the diverging lens = its focal length.
=> L-fc = -fd
=> L = fc -fd = distance between the lenses
ii) If parallel rays enter the diverging lens first then the object distance for it = p = infinity
image distance = fd
Distance between the lenses = L
object distance for the convex lens = L-fd
If the final rays are to be parallel image is formed at the focal point of the convex lens and inbetween the two lenses.
So we cannot have the same outcome.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.