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Formula for a concave mirror...

  • 30-09-2012 03:56PM
    #1
    graate
    Registered Users, Registered Users 2 Posts: 15 ✭


    Hey everyone,

    I'm just starting light (leaving cert) and my maths is really far behind and I'm struggling (to my shame) doing a bit of work that involves basic fractions...

    Basically the question is...

    An object is placed 12cm in front of a concave mirror of a focal length 20cm. Find the position, nature and magnification of the image. Let us assume we do not know whether the image is real or virtual and use the formula 1/u + 1/v = 1/f.


    So the solution is

    1/12 + 1/v = 1/20

    1/v = 1/20 - 1/12

    1/v = 3-5/60

    =-1/30

    v = -30cm


    Now I understand everything except for the part in bold. I really have no idea how 1/20 - 1/12 = 3-5/60. Could somebody please save my life and explain this to me?

    Physics once again showing me how badly I need to go over my basic maths again....

    Thanks!


Comments

  • #2
    gctest50
    Registered Users, Registered Users 2 Posts: 9,534 ✭✭✭gctest50


    Common denominator ? see little jpg attached


  • #3
    krd
    Banned (with Prison Access) Posts: 3,449 ✭✭✭krd


    graate wrote: »
    Now I understand everything except for the part in bold. I really have no idea how 1/20 - 1/12 = 3-5/60. Could somebody please save my life and explain this to me?

    Your fractions are correct. I couldn't tell you about the rest - I hated lens formulas and exercises with a passion.
    Physics once again showing me how badly I need to go over my basic maths again....

    Practice your basic maths to death. Little basic maths flubs will cost you very heavily.

    The part of the department of education who do the exams have a website. They have the past examination papers, plus they have the sheets used by the examiners to mark the paper.


  • #4
    Turtwig
    Registered Users, Registered Users 2 Posts: 26,578 ✭✭✭✭Turtwig


    Let's try to write the fractions a different way.
    [latex]
    \frac{1}{20} -\frac{1}{12} = \frac{1*(20)*(12)}{20*(20)*(12)}-\frac{1*(20)*(12)}{12*(20)*(12)}
    [/latex]
    All I've done here is multiply both fractions by 1. [latex]\left(\frac{(20)*(12)}{(20)*(12)\right)}[/latex]. So essentially I've done nothing but expressed the same quantity in a different way. Cancelling out the commons, leaves you with:
    [latex]
    \frac{1*12}{20*12}-\frac{1*20}{12*20}\Rightarrow \frac{1}{20}-\frac{1}{12}
    [/latex]

    Can you see anyway I could change the above step to get the [latex]\frac{3}{60}-\frac{5}{60}[/latex]

    Hint : What if instead of cancelling the matching pairs 20/20 and 12/12 you tried to cancel the non matching ones? [latex]e.g\left(\frac{12}{20}\;and\; \frac{20}{12}\right)[/latex]


  • #5
    graate
    Registered Users, Registered Users 2 Posts: 15 ✭graate


    Sorry forgot to thank you guys! All help is appreciated more than you know.


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