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    Basically, this is an optical fibre with radius b. The refractive indeces for the fibre and the cladding are n2 and n3, respectively, where n3 < n2. In the diagram above, T = theta. EDIT: Ooh also - T3 = theta_c!

    I need to show that

    \sin \theta_1 = \sqrt{n_{2}^{2} - n_{3}^{2}}.

    First I considered some of the relations between critical angles (theta_c), angles of incidence, etc:

    n_1 \sin \theta_1 = n_2 \sin \theta_2

    sin \theta_c = \frac{n_3}{n_2} = \frac{a}{c}

    sin \theta_2 = \frac{b}{c}

    and b = \sqrt{c^2 - a^2}.

    My thoughts are that I need to show that sin\theta_1 = b, c = n2 and a = n3. I wasn't really sure how to go about this though, and I feel like I'm missing some obvious geometric/trigonometric relationships here.

    Any tips?
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    Uhhh, I think I just did it :|

    \sin \theta_c = a/c = n_3 / n_2

    but \cos\theta_2 = a/c = \sin\theta_c

    And \cos^2 \theta_2 = 1 - \sin^2 \theta_2

    But \sin\theta_2 = (1/n_2) \sin\theta_1 as n1 = 1 for air

    So  \cos^2 \theta_2 = 1 - \sin^2 \theta_2 = 1- (1/n_{2}^{2}) \sin^2 \theta_1 = a^2 / c^2

    Which reduces to:

    n_{2}^{2} - \sin^2 \theta_1 = n_{2}^{2} (a^2 / c^2)

    but a^2 / c^2 = n_{3}^{2} / n_{2}^{2}

    So \sin^2 \theta_1 = n_{2}^{2} - n_{3}^{2} \Rightarrow \sin \theta_1 = \sqrt{n_{2}^{2} - n_{3}^{2}}
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    (Original post by trm90)


    Basically, this is an optical fibre with radius b. The refractive indeces for the fibre and the cladding are n2 and n3, respectively, where n3 < n2. In the diagram above, T = theta. EDIT: Ooh also - T3 = theta_c!

    I need to show that

    \sin \theta_1 = \sqrt{n_{2}^{2} - n_{3}^{2}}.

    First I considered some of the relations between critical angles (theta_c), angles of incidence, etc:

    n_1 \sin \theta_1 = n_2 \sin \theta_2

    sin \theta_c = \frac{n_3}{n_2} = \frac{a}{c}

    sin \theta_2 = \frac{b}{c}

    and b = \sqrt{c^2 - a^2}.

    My thoughts are that I need to show that sin\theta_1 = b, c = n2 and a = n3. I wasn't really sure how to go about this though, and I feel like I'm missing some obvious geometric/trigonometric relationships here.

    Any tips?
    hey,
    dont you also need to consider what happens when it refracts from the fibre to cladding?Assuming TIR occurs , can we not say n_2sin\theta_3\

=n_3sin90..
    also sin\theta_2=cos\theta_3


    further hint
    Also note from the refractive index of air=1
    so the first equation becomes \sin \theta_1 = n_2 \sin \theta_2
    but we know sin\theta_2=cos\theta_3 from the triangle
    subbing this in and rearrnaging for costheta_3
    we get cos\theta_3 =

\frac{sin\theta_1}{n_2}
    also from above sin\theta_3=\frac{n_3}{n_2}
    now using sin^2\theta_3+cos^2\theta_3=1 gives the desired result


    Edit ;too late :sigh:
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    (Original post by rbnphlp)
    hey,
    dont you also need to consider what happens when it refracts from the fibre to cladding?Assuming TIR occurs , can we not say n_2sin\theta_3\

=n_3sin90..
    also sin\theta_2=cos\theta_3


    further hint
    Also note from the refractive index of air=1
    so the first equation becomes \sin \theta_1 = n_2 \sin \theta_2
    but we know sin\theta_2=cos\theta_3 from the triangle
    subbing this in and rearrnaging for costheta_3
    we get cos\theta_3 =

\frac{sin\theta_1}{n_2}
    also from above sin\theta_3=\frac{n_3}{n_2}
    now using sin^2\theta_3+cos^2\theta_3=1 gives the desired result


    Edit ;too late :sigh:
    Thanks anyway !

    (I feel a bit silly now - about 50% of the physics threads I've posted on TSR have been useless cause I always solve it five minutes later. I think it's cause typing it out relaxes me and helps me think about the question more slowly and clearly)
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    (Original post by trm90)
    Thanks anyway !

    (I feel a bit silly now - about 50% of the physics threads I've posted on TSR have been useless cause I always solve it five minutes later. I think it's cause typing it out relaxes me and helps me think about the question more slowly and clearly)
    no worries..it happens to me a lot with my maths threads:p: ...
 
 
 
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