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Thread: how do i find this?

  1. #1
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    how do i find this?

    how do I figure this out, If someone can please guide me in steps as im struggling.

    the question is: An object is 3cm high is placed 40cm in front of a thin +10.00D lens. Find the position and size of the image.

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    really? no help. whats the point of this website

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    londongirl this is a specific physic problem. I remember doing problems like these. Here is a similar problem that might help explain how to solve this. I dont know if this is what your looking for, but it might help. I would just google a similar question and see what you come up with. I don't remember too many opticians that can recall lots of physics on the spot.

    Question: An object is located 25 cm in front of a +5.00D lens.

    1. What is the vergence of the incident rays?
    Use the following equation to calculate the object vergence: U = 100/u
    Where u = object location in cmU = object vergence = 100/–25 = –4.00D

    2. What is the refracting vergence?
    Use the equation U + D = V where
    U = object vergence = –4.00D
    D = lens power = +5.00D
    V = refracting vergence of the image = –4.00D + (+5.00D) = +1.00D

    3. Where is the image located?
    Use the equation, v = image position = 100/V where V = +1.00D
    v = image position = 100/+1 = +100 cm to the right of the lens.

    4. Is the image real or virtual?
    Real

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    Quote Originally Posted by bta89 View Post
    Question: An object is located 25 cm in front of a +5.00D lens.

    1. What is the vergence of the incident rays?
    Use the following equation to calculate the object vergence: U = 100/u
    Where u = object location in cmU = object vergence = 100/–25 = –4.00D

    2. What is the refracting vergence?
    Use the equation U + D = V where
    U = object vergence = –4.00D
    D = lens power = +5.00D
    V = refracting vergence of the image = –4.00D + (+5.00D) = +1.00D

    3. Where is the image located?
    Use the equation, v = image position = 100/V where V = +1.00D
    v = image position = 100/+1 = +100 cm to the right of the lens.
    This is correct as far as it goes and is a simple example of vergence through a ‘thin’ lens. To answer the size of the image question, you will need a working knowledge of basic ray tracing rules. Then you will see the relationship of object height and image height, and the similar triangles produced by ray tracing through this thin lens...
    object height/object distance = image height/image distance
    3cm/40cm = xcm/100cm
    300/40 = x
    7.5cm = x = height of image
    But in the real world there is no such thing as a ‘thin’ (0 thickness) lens. Ray-tracing and Vergence through a ‘thick’ lens with its index of refraction, centre thickness and front/back curves is much more complicated. The ‘thick’ lens contains primary and secondary principal planes with corresponding nodal points at the lens axis, and you must ray trace through these and amend your ‘similar triangles’ numbers accordingly. And using other more-complicated ‘vergence’ formulas, you will see the shift of these principal planes towards the stronger of the two lens curves, resulting in different image size for the same lens power. This is why an increase in the front curve (base curve) and/or centre thickness of a given lens power can change the size of image, and understanding this can be useful in a troubleshooting situation or even designing iseikonic lenses.
    A formal optical education with qualified instructors is so very important to develop one into a fully-qualified optician. I recall one student who stated after completion of ‘Vergence’ that he wanted to design a t-shirt emblazed with “I Survived Vergence!!”
    Last edited by tmorse; 02-06-2019 at 06:37 PM.

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    What's up? drk's Avatar
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    It seems we have a lot of "new opticians" from England with optics questions on here.

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    Master OptiBoarder optical24/7's Avatar
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    Quote Originally Posted by drk View Post
    It seems we have a lot of "new opticians" from England with optics questions on here.
    And also impatient and non-comprehending “The point of this website “. Hint...There are knowledgeable folks here, but they are not here solely to help you with your homework.
    Last edited by optical24/7; 02-06-2019 at 07:18 PM.

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    Sorry, but I have made an error in my earlier post by using some of bta89’s question numbers rather than your original question numbers.

    QUESTION: 3cm object sitting 40cm in front of a +10.00 lens.

    Object distance is noted as small l and incoming vergence L (in diopters) = n(air)/l = 1/object distance l(in m) or 100/object distance l(in cm). Object rays are in front of lens and are diverging before striking lens, so they will have a minus vergence.

    So L = 100/-l(cm) = 100/-40cm = -2.50D vergence power of object striking this +10.00 lens.
    L (incoming object vergence) + F(lens power) = L’ (exiting vergence of image) in dioptres
    -2.50 + (+10.00) = +7.50D = L’
    and image distance l’ is reciprocal of this exiting vergence L’ . So L’ = 1/l’(m) or 100/l’(cm).
    Thus l’ = 100/L’, so image distance l’ = 100/(+7.50) = 13.3cm to right of lens.
    Ray tracing object through this +10.00 lens produces similar right-angle triangles. Thus
    Height of object/object distance = Height of Image/image distance
    3cm/40cm = x/13.3cm
    40cm/40cm = x, so image height is 1cm.
    Hope this now make better sense.
    Last edited by tmorse; 02-08-2019 at 04:54 PM.

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