Discussion Forum

 Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n₁  surrounded by a medium of lower refractive index n₂. The light guidance in  the structure takes place due to successive total internal reflections at the interface of the media n₁  and n₂ as shown in the figure. All rays with the angle of incidence i less than a particular value i_m are confined in the medium of refractive index n₁ . The numerical aperture  (NA) of the structure is defined as sin i_m.

If two structures of same cross-sectional area, but different numerical  apertures NA₁  and NA₂  (NA₂ <NA₁) are joined  longitudinally, the numerical aperture  of the combined structure is 

Answer:

Explanation:

       $\sin i_{m}=n_{1}\sin (90-\theta_{c})$ 

$\Rightarrow$   $\sin i_{m}=n_{1}\cos\theta_{c}$

$\Rightarrow$                               $NA=n_{1}\sqrt{1-\sin^{2}\theta_{c}}$

                                            $NA=n_{1}\sqrt{1-\frac{n_{2}^{2}}{n_{1}^{2}}}=\sqrt{n^{2}_{1}-n^{2}_{2}}$

 Substituting the values we get,

       $NA_{1}=\frac{3}{4}$   and   $NA_{2}=\frac{\sqrt{15}}{5}=\sqrt{\frac{3}{4}}$

  NA₂  <NA₁  

   Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical  aperture. Which is NA₂