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Physics Lenses Question watch

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    http://www.aqa.org.uk/qual/pdf/AQA-4451-W-SM-07.PDF

    Page 187, part a)

    This is the one little topic in physics which I am clueless on. I've read my books but they just seem to drone on about how you should be able to do it, but not actually tell you how to?

    Need advice.
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    I don't understand either. I can do topics 1 and 4, not 2 or 3. If I somehow manage to get an A it's a miracle.
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    Ok, I may be wrong but you might have to:
    Draw a horizontal line from the top of the object to the axis thing
    Draw a diagonal line from there to the focus
    Draw another diagonal line going through the centre. ie. where the lense is so (0,0)
    See where they meet up by tracing the lines back

    Does that give you anything?
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    (Original post by piece_by_piece)
    Ok, I may be wrong but you might have to:
    Draw a horizontal line from the top of the object to the axis thing
    Draw a diagonal line from there to the focus
    Draw another diagonal line going through the centre. ie. where the lense is so (0,0)
    See where they meet up

    Does that give you anything?
    I thought of just that - something like that's in the revision guide I have here, but anyway, why would the line from the top of the object refract from the vertical axis to the focal point? It's just empty air, nothing is there, the lens is tiny :confused:
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    (Original post by Torseor)
    I thought of just that - something like that's in the revision guide I have here, but anyway, why would the line from the top of the object refract from the vertical axis to the focal point? It's just empty air, nothing is there, the lens is tiny :confused:
    I have no idea, that's how we've been taught to do them
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    SEE ATTACHED PICTURE FOR ANSWER

    light ray parallel to principal axis is refracted through principal focus. And light ray passing through centre of lens continues straight through (the red lines). As it is a virtual image (as object is under one focal length away from lens) you have to track light rays back (the orange lines) to form an magnified, upright, virtual image (big black arrow)

    hope that helps...

    edit: the confusing thing with this question is you have to ignore the fact that the lens is tiny and just imagine it goes all the way to the top.

    If the lens wasn't closer than 1 focal length, you can draw a third line in to check your accuracy and this goes through the focal point on the obkect's side of the lens and then comes out parallel (it is just the opposite of the first lens in this example)
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    Why would it be virtual? What is the difference between virtual and real anyway, and how can I tell?!
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    (Original post by piece_by_piece)
    Why would it be virtual? What is the difference between virtual and real anyway, and how can I tell?!
    because the object is between the focalpoint (F) of the lens and the lens itself. If you look at the rays after they have passed through the lens, they are diverging (i.e spreading out); therefore, they will never meet at a point --> image must be virtual

    real image = can be captured on a screen where the light rays actually converge (e.g. image on eye, or cinema screen)
    virtual image = cannot be captured on a screen and does not actually 'exist'. The eye looks back through the lens and sees the virtual rays (the orange ones) converging
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    (Original post by hughey)
    because the object is between the focalpoint (F) of the lens and the lens itself. If you look at the rays after they have passed through the lens, they are diverging (i.e spreading out); therefore, they will never meet at a point --> image must be virtual

    real image = can be captured on a screen where the light rays actually converge (e.g. image on eye, or cinema screen)
    virtual image = cannot be captured on a screen and does not actually 'exist'. The eye looks back through the lens and sees the virtual rays (the orange ones) converging
    Thanks So if I have to track the lines back then it's a virtual image?
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    Sweet hughey, thanks. I think the key point to remember is that it doesn't matter how big the lens is.. ?
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    (Original post by piece_by_piece)
    Thanks So if I have to track the lines back then it's a virtual image?
    You can just see whether it's a virtual image because the actual image is "behind" the mirror, which is not "real", so on hughey's diagram, if the lines crossed right in front of the eye it would be real.


    I think. This is the most confusing topic across all GCSEs I've done, by quite a long shot :confused:


    Converging + far = real
    Converging + near = virtual
    Diverging = always virtual

    Concave mirror + far = real
    Concave mirror + near = virtual
    Convex mirror = always virtual

    I hope that's right :eek:
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    thanks for the rep whoever it was.

    i think generally if they give you a picture of the eye and say how the eye forms the image, then it will be a virtual image.

    any tracking back is virtual

    btw, that question is the same as a magnifying glass.

    as it appears to be confusing, i have put together some NOTES AND DIAGRAMS ON THE LENSES AND MIRRORS stuff - see attached

    hope that helps...

    p.s. the lens stuff is in two documents; i apologise if the pics are a bit small - it was the only way i could get it to be below the upload limit.
 
 
 

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