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Formula of refraction at spherical surface

  1. Aug 15, 2014 #1
    1. The problem statement, all variables and given/known data
    i am confused which eqaution to use for formula of refraction at spherical surface , do i need to put a modulus for n2-n1 ? some book gives modulus , while the other book not . which one is correct?

    gy5uYRJ.gif
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 15, 2014 #2

    ehild

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  4. Aug 16, 2014 #3
  5. Aug 16, 2014 #4

    ehild

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    I might have misunderstood you. What do you mean on "modulus"? Absolute value?

    Check the sign convention of R and that of the image distance before using the formula. It might differ from book to book.

    ehild
     
  6. Aug 16, 2014 #5
    yes , absolute value , i have an example here.
    assuming the light ray pass from left to right, since the refracting surface is concave to the incident light ray, then r is NEGATIVE , but if I put modulus to n2-n1 , my ans would be diffrerent from the text, tat's why i am confused whether to use modulus for n2-n1 or not . in the text , it can be known that n2-n1 is negative.
     

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  7. Aug 16, 2014 #6

    ehild

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    The solution uses the formula ##\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{R}## (this is more common than the one you used) with the sign convention: R is positive if the spherical surface is convex as viewed by the incoming ray. R is negative if the surface is concave from the direction the light arrives. The image distance u is positive if the image is at that opposite from where the light arrives, and negative if it is at the same side. The object distance is positive if it is in front of the surface, ( at that side from where the light arrives).

    The sign convention in case of the formula you cited is that the distances to the left from the surface are negative and they are positive if they are to the right of the surface. In that case u would be -30 cm.

    In case of the problem you show, R is negative, R=-20 cm. n1=1, n2=1.5, n2-n1=0.5. If you calculate the image distance with the formula ##\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{R}##, ##\frac{1}{u}+\frac{1.5}{v}=\frac{0.5}{-20}## you get the result in the book.

    Using your previous formula, both the object and the centre of the sphere are to the left from the refracting surface, so both u and R are negative. u=-30 cm, R=-20 cm, so you get ##\frac{1.5}{v}-\frac{1}{-30}=\frac{0.5}{-20}## which gives the same result.

    Do not put absolute value anywhere.

    ehild
     
  8. Aug 16, 2014 #7
    i still cant understand why the u is -30cm. the light ray passed from left to right, (the object is placed to the left of the refracting surface) , so the object is REAL am i right? so it should be POSITIVE ?
     
  9. Aug 16, 2014 #8
    here's another case . now , the n2-n1 is negative where 1-1.5 = -0.5

    if i put modulus on it , my n2-n1 is always positive , but the book didnt put modulus on it. here's a better case showing whether the n2-n1 should be placed modulus or not.
     

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  10. Aug 16, 2014 #9

    ehild

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    If you use the sign convention that the real object has positive distance, then apply the formula ##\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{R}##. The other one you cited in your first post is valid in case of the other sign convention.


    ehild
     
  11. Aug 16, 2014 #10

    ehild

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    Forget the modulus. u=3 cm, R=-10 cm, n1=1.5 and n2=1. Plug in the data into the formula ##\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{R}##


    ehild
     
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