LA.L Combination lens. Two converging lens are put together in combination with one another. They are separated by 7.0 meters and placed so that the focus point of the first lens is at the same place as the focus point of the second lens. If the first lens has a focus of fi = 2.5 meters, answer the following questions if an object with a height of 1.25 m is placed in front of the first lens at a distance of 6.0 meters: (a) [1 pt] What are pi, 41, f1, and m?? (b) 3 pts) What are p2, 42, fa, and ma? (c) [1 pt) Determine Mtotal and the final image height, Htotal - (d) (3 pts] Draw a ray-tracing diagram on the graph paper.
Solution
1 a)Since the focal point of the first lens is at the same place as the focal point of the second lens, it is obvious that that the focal length of the second lens f 2 =7-2.5=3.5 m.
The object distance for lens1 p 1 =6 m.
We have the lens equation for lens 1:
1/p 1 +1/q 1 =1/f 1 i.e. 1/6+1/q 1 =1/2.5
Therefore 1/q 1 =0.23 i.e q 1 =4.35 m.
f 1 =2.5 m. m 1 =-q 1 /p 1 =-4.35/6=-0.73.
b)The image formed by lens 1 is the object for lens 2. Therefore the object distance for lens2 p 2 =7-4.35=2.65 m.
The focal length of lens 2 f 2 =3.5 m.
We have 1/p 2 +1/q 2 =1/f 2 i.e 1/2.65+1/q 2 =1/3.5
Therefore 1/q 2 =1/3.5-1/2.65=0.29-0.38=-0.09 i.e q 2 =-1/0.09=-11.1 m.
The final image is formed on the left side of lens1 at a distance of 11.1-7=4.1 m from it.
The magnification of lens 2,m 2 =-q 2 /p 2 =-11.1/2.65=-4.2.
c) The total magnification due to both the lenses M total =m 1 *m 2 =-0.73*-4.2=3.05.
The final image height h total =M total *h 1 =3.05*1.25=3.8 m.
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LA-L Combination lens- Two converging lens are put together in combina.docx
1. LA.L Combination lens. Two converging lens are put together in combination with one another.
They are separated by 7.0 meters and placed so that the focus point of the first lens is at the same
place as the focus point of the second lens. If the first lens has a focus of fi = 2.5 meters, answer
the following questions if an object with a height of 1.25 m is placed in front of the first lens at a
distance of 6.0 meters: (a) [1 pt] What are pi, 41, f1, and m?? (b) 3 pts) What are p2, 42, fa, and
ma? (c) [1 pt) Determine Mtotal and the final image height, Htotal - (d) (3 pts] Draw a ray-
tracing diagram on the graph paper.
Solution
1 a)Since the focal point of the first lens is at the same place as the focal point of the second lens,
it is obvious that that the focal length of the second lens f 2 =7-2.5=3.5 m.
The object distance for lens1 p 1 =6 m.
We have the lens equation for lens 1:
1/p 1 +1/q 1 =1/f 1 i.e. 1/6+1/q 1 =1/2.5
Therefore 1/q 1 =0.23 i.e q 1 =4.35 m.
f 1 =2.5 m. m 1 =-q 1 /p 1 =-4.35/6=-0.73.
b)The image formed by lens 1 is the object for lens 2. Therefore the object distance for lens2 p 2
=7-4.35=2.65 m.
The focal length of lens 2 f 2 =3.5 m.
We have 1/p 2 +1/q 2 =1/f 2 i.e 1/2.65+1/q 2 =1/3.5
Therefore 1/q 2 =1/3.5-1/2.65=0.29-0.38=-0.09 i.e q 2 =-1/0.09=-11.1 m.
The final image is formed on the left side of lens1 at a distance of 11.1-7=4.1 m from it.
The magnification of lens 2,m 2 =-q 2 /p 2 =-11.1/2.65=-4.2.
2. c) The total magnification due to both the lenses M total =m 1 *m 2 =-0.73*-4.2=3.05.
The final image height h total =M total *h 1 =3.05*1.25=3.8 m.
![LA.L Combination lens. Two converging lens are put together in combination with one another.
They are separated by 7.0 meters and placed so that the focus point of the first lens is at the same
place as the focus point of the second lens. If the first lens has a focus of fi = 2.5 meters, answer
the following questions if an object with a height of 1.25 m is placed in front of the first lens at a
distance of 6.0 meters: (a) [1 pt] What are pi, 41, f1, and m?? (b) 3 pts) What are p2, 42, fa, and
ma? (c) [1 pt) Determine Mtotal and the final image height, Htotal - (d) (3 pts] Draw a ray-
tracing diagram on the graph paper.
Solution
1 a)Since the focal point of the first lens is at the same place as the focal point of the second lens,
it is obvious that that the focal length of the second lens f 2 =7-2.5=3.5 m.
The object distance for lens1 p 1 =6 m.
We have the lens equation for lens 1:
1/p 1 +1/q 1 =1/f 1 i.e. 1/6+1/q 1 =1/2.5
Therefore 1/q 1 =0.23 i.e q 1 =4.35 m.
f 1 =2.5 m. m 1 =-q 1 /p 1 =-4.35/6=-0.73.
b)The image formed by lens 1 is the object for lens 2. Therefore the object distance for lens2 p 2
=7-4.35=2.65 m.
The focal length of lens 2 f 2 =3.5 m.
We have 1/p 2 +1/q 2 =1/f 2 i.e 1/2.65+1/q 2 =1/3.5
Therefore 1/q 2 =1/3.5-1/2.65=0.29-0.38=-0.09 i.e q 2 =-1/0.09=-11.1 m.
The final image is formed on the left side of lens1 at a distance of 11.1-7=4.1 m from it.
The magnification of lens 2,m 2 =-q 2 /p 2 =-11.1/2.65=-4.2.](https://image.slidesharecdn.com/la-lcombinationlens-twoconverginglensareputtogetherincombina-230204022000-e2d11a05/85/lal-combination-lens-two-converging-lens-are-put-together-in-combinadocx-1-320.jpg)
