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Homework Help: Dielectric constant of a rod

  1. Dec 28, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-12-28_12-27-28.png

    2. Relevant equations


    3. The attempt at a solution
    ## n_1 = \sqrt{ \epsilon _1} ## .....(1) ,

    ## n_2 = 1 ## .....(2)

    ## \frac { \sin {\theta_i}}{ \sin {\theta_l} } = \frac { n_1}{n_2} = n_1 ## .....(3)

    ## \cos{\theta_1} = \frac { n_2}{n_1} = \frac1{ n_1} ## .....(4)

    According to the question, the dielectric constant should be such that even when the incident angle is slightly less than ## \frac { \pi} 2 ## , the ray should come out of the rod without getting absorbed.

    So, ## \theta _i \leq \frac { \pi} 2 ## .....(5)

    Taking ##\theta _i = \frac { \pi} 2 ## , .....(6) I don't know why I am taking this.


    ## \frac1{ \sin {\theta_l} }= n_1 ## .....(7)

    ## \cos{\theta_1} = \frac1{ n_1} ## .....(8)

    From (7) and (8), I get

    ## n_1 = \sqrt{ 2} ## .....(9)

    From (1), ## \epsilon _1 = n_1^2 ## .....(10)

    So, that dielectric constant = 2, option (c).

    Is this correct?
     

    Attached Files:

  2. jcsd
  3. Dec 28, 2017 #2

    Charles Link

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    Snell's law is written ## n_1 \sin(\theta_1)=n_2 \sin(\theta_2) ##. For this case ## n_1=\sqrt{\epsilon_{r1}} ##, and ## n_2 =1 ##. ## \\ ## ## \theta_1 ## is the angle of incidence inside the material, measured from the normal. There is a ## \theta_1 ## for which the left side of the equation is equal to 1. What happens if ## \theta_1 ## is such that the left side of the equation is greater than 1 ? Can you get a solution for the right side in that case, to determine the emerging angle of the refracted ray? ## \\ ## Editing: This problem is slightly tricky: If ## \theta_i=90^o ##,(the steepest angle of incidence at the entry point), ## \theta_r=\sin^{-1}(1/n_1) ##. The resulting ## \theta_1=90^o-\theta_r ##. I'll let you try to finish it up. Meanwhile, the file you uploaded is apparently the wrong one.
     
    Last edited: Dec 28, 2017
  4. Dec 28, 2017 #3

    haruspex

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    As far as I can make out, this matches what @Pushoam has done. Have you identified an error in the working?
     
  5. Dec 28, 2017 #4

    Charles Link

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    The OP's statement after equation (6) was somewhat confusing. Meanwhile, I was expecting to see the statement ## n_1 \sin(\theta_1)>1 ## for total internal reflection. In addition, ## \sin(\theta_1)=\cos(\theta_r)=\frac{\sqrt{n_1^2-1}}{n_1} ##. Thereby ## \sqrt{n_1^2-1}>1 ##, so that ## n_1^2=\epsilon_{r1}>2 ##. ## \\ ## I didn't see these details in the OP's solution. ## \\ ## (It was difficult to answer the question by @haruspex without providing the solution.) ## \\ ## From what I can see, the OP gets the right answer, but the algebraic steps to the answer are not readily apparent. Perhaps the OP performed a similar algebra, but too many steps were omitted to readily tell how the OP arrived at the answer. :) ## \\ ## In any case, I think the OP @Pushoam might find these details useful.
     
    Last edited: Dec 28, 2017
  6. Dec 28, 2017 #5

    haruspex

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    Ok, thanks for clarifying.
     
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