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Geometrical optics- position of image

  1. Aug 11, 2014 #1
    1. The problem statement, all variables and given/known data

    A sphere of radius 12.0cm has refractive index of 1.33. A speck is 4.0cm from the centre of the sphere is viewed along the diameter that passes thru the speck. Find the position of image when the speck is viewed from the nearer side?
    the ans is -7.2cm

    2. Relevant equations



    3. The attempt at a solution
    n1= 1.33 n2= 1 , u = +0.08m. Assuming light travel form right to left , The nearer position from the speck to the left refracting surface is +0.08m. so this lead me to (1.33/0.08) +1/(v) = (1-0.33)/(-0.12) , my v is -0.05m . r is negative because the refracting surface is concave to the object. i just cant understand why I am wrong ? please help! p/s : the speck is inside the spherical (ball) am i right?
     
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  3. Aug 11, 2014 #2

    Simon Bridge

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    Last edited: Aug 11, 2014
  4. Aug 12, 2014 #3
    this is what i did exactly ..anything wrong with it?
     
  5. Aug 12, 2014 #4
    sorry it should be (1.33/0.08) +1/(v) = (1.33-0.33)/(-0.12) where my v is -0.05m . what's wrong with it?
     
  6. Aug 12, 2014 #5

    Simon Bridge

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    ... so it was a typo - as you wrote it out it was wrong as you noticed.

    Did you try verifying your result sing a ray diagram?
    Did you check what you did against the link I gave you?
     
  7. Aug 12, 2014 #6
    I checked thru the link. But I stillcant figure out which part I did wrongly.
     
  8. Aug 12, 2014 #7

    Simon Bridge

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    Did you try verifying your result using a ray diagram?
    Basically your method is correct but you look like you keep making mistakes in substitution.

    ... which one is n1 and which n2?

    I'm getting different values of v to you for the same calculations.
    Show me the arithmetic one step at a time.
     
    Last edited: Aug 12, 2014
  9. Aug 12, 2014 #8
    n1= 1.33 , n2 = 1.00 , so n1-n2= 1.33-1.00
     
  10. Aug 12, 2014 #9

    Simon Bridge

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    In the equation you are using, is that n1-n2 in the denominator, or n2-n1?
    The equation is also at the bottom of that webpage I gave you earlier.
     
  11. Aug 14, 2014 #10

    i am using modulus of n2-n1, as in the book. is there's anything wrong with the equation?
     

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  12. Aug 14, 2014 #11

    Simon Bridge

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    ... well, only that you just told me that you did:
    ...

    But you told me that n1=1.33 and n2=1 (post #1) ... so where did that 0.33 come from, and why is n1 first?
    What you wrote there does not match your notes, and nor does it match the link I gave you.
    surely: n2 - n1 = 1 - 1.33?

    Try doing the algebra one line at a time. Be super careful: you appear to have a blind spot there.
     
  13. Aug 14, 2014 #12
    n1 = 1.33 , n2= 1.00 , u = +0.08 , n1-n2 = 1.33-1.00=0.33 , r = -0.12
    (1.33/0.08) + (1/v)= (0.33/-0.12) , the v i get is still -0.0516m . i cant figure out which part i did wrongly
     
  14. Aug 14, 2014 #13
    but suprisingly if I take r = +0.12 , i got the ans which is 0.072m . by taking r= -0.12 , i assume the refreacting surface is concave relative to the incident light ray.
     
  15. Aug 15, 2014 #14

    Simon Bridge

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    We don't normally do the problem for people, but you are stuck on the substitution and you have sincerely tried hard.
    ... the bold face is incorrect.
    What is supposed to go in there is n2-n1 and you wrote down n1-n2.
    It is very important to write down the equation you are using with all symbols before you do any substituting.

    Using:$$n_1=1.33,\; n_2=1,\; o=8\text{cm},\; v=\text{?},\; R=-12\text{cm}$$
    $$\frac{n_1}{o}+\frac{n_2}{v}=\frac{n_2-n_1}{R}\\
    \implies \frac{1.33}{8}+\frac{1}{v} = \frac{1-1.33}{-12} = \frac{0.33}{12}\\
    \implies v= \left( \frac{0.33}{12}- \frac{1.33}{8} \right)^{-1} = -7.21\text{cm}$$ ... now do you see what happened?
     
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