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Newton's Ring Apparatus and Refractive Index

  1. Apr 2, 2005 #1
    Hello, I'm a bit stumped on a problem and wondered if anyone knew how to approach this problem:

    When a Newton's ring apparatus, ( Fig. 24-30(see attached) ) is immersed in a liquid, the diameter of the eighth dark ring decreases from 2.99 cm to 2.49 cm. What is the refractive index of the liquid?

    Any ideas would be great, thanks.

    Attached Files:

    Last edited: Apr 2, 2005
  2. jcsd
  3. Apr 2, 2005 #2


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    Homework Helper

    This may apply

    [tex]r= \sqrt(m \lambda R/n) [/tex]

    substitute and solve for n
  4. Apr 2, 2005 #3
    I am not sure that I can do that, the problem posted above is all the information I am given. I do not know the wavelength. (also, what does the R represent?) Is there something I missed?

    thank you
  5. Apr 3, 2005 #4


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    Staff: Mentor

    Use the equation twice:

    Initially, the apparatus is in air (presumably). [itex]n = 1[/itex] and [itex]r = 2.99[/itex] cm.

    When immersed in the liquid, [itex]n = unknown[/itex] and [itex]r = 2.49[/itex] cm.

    [itex]R[/itex] is the radius of the spherical surface that is part of the Newton's Rings apparatus. You don't need to know its value, nor do you need to know the wavelength. When you set up the two equations above, you should be able to figure out why. :smile:
  6. Apr 3, 2005 #5
    Thank you, I got the correct answer by using the equation above. However, I do not know where the equation above came from. I would like to know why the equation above works. Could you tell me how you came up with the equation please? In class we did not cover the radius of the spherical surface of Newton's apparatus, so needless to say, I have no equations that involve anything like it. Again, thank you for your help, I understand how to manipulate the above equation but I don't know why the above equation is valid. Any help if you are so inclined. Thanks.
  7. Apr 3, 2005 #6


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    Staff: Mentor

    Searching for "Newton's rings" with Google gives this as the very first result:


    It ends up with [itex]m+1/2[/itex] in the equation instead of plain [itex]m[/itex], but that's probably because one is for bright rings (constructive interference) and the other is for dark rings (destructive interference).
  8. Apr 3, 2005 #7
    wow, i feel kinda foolish. i'm usually much better at being thorough with looking for something online. sorry but again, thank you for your help. i'm glad that i understand it now, thanks.
  9. Dec 14, 2010 #8
    srry know its too late to answer, refractive index=(2.99^2)/(2.49^2)=1.44
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