Concept explainers
Two lenses, one converging with focal length 20.0 cm and one diverging with focal length -10.0 cm. are placed 25.0 cm apart. An object is placed 60.0 cm in front of the converging lens. Determine (a) the position and (b) the magnification of the final image formed, (c) Sketch a ray diagram for this system.
(a)
The position of the image.
Answer to Problem 64P
Solution:
The image forms in the right of second lens.
Explanation of Solution
Given:
Converging lens has focal length and the diverging one has focal length . They are placed apart. Object is placed before the first lens.
Formula used:
Lens equation,
and are lens to focal, lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.
Magnification,
Calculation:
cm
cm
Image distance formed by first lens
Using lens equation,
This gives, cm
Now, the distance between two lenses is 25 cm.
The image formed by first lens acts as a virtual object for second lens.
Thus, object distance for second lens is cm
Considering final image distance from second lens as we get,
Focal length of the diverging lens cm
To find the position substitute the values to lens equation, we get,
(b)
Magnification of the image formed by two lenses, one converging with focal length and one diverging with focal length placed apart with object placed in front of converging lens.
Answer to Problem 64P
Solution:
The magnification is .
Explanation of Solution
Given:
Converging lens has focal length and the diverging one has focal length . They are placed apart. Object is placed before the first lens.
Formula used:
Lens equation,
and are lens to focal, lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.
Magnification,
Calculation:
cm
cm
Image distance formed by first lens
Using lens equation,
This gives, cm
Now, the distance between two lenses is 25 cm.
The image formed by first lens acts as a virtual object for second lens.
Thus, object distance for second lens is cm
Considering final image distance from second lens as we get,
Focal length of the diverging lens cm
Thus, the total magnification is =1.
(c)
The ray diagram of the system.
Answer to Problem 64P
Solution:
Explanation of Solution
Given:
Converging lens has focal length and the diverging one has focal length . They are placed apart. Object is placed before the first lens.
Formula used:
Lens equation,
and are lens to focal, lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.
Magnification,
Calculation:
cm
cm
Image distance formed by first lens
Using lens equation,
This gives, cm
Now, the distance between two lenses is 25 cm.
The image formed by first lens acts as a virtual object for second lens.
Thus, object distance for second lens is cm
Considering final image distance from second lens as we get,
Focal length of the diverging lens cm
Ray diagram of the system is as below-
Chapter 23 Solutions
Physics: Principles with Applications
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