Derive an expression for numerical aperture of a step index optical fibre. Give its physical significance.

Set-up: A step-index fibre has a core of index \(n_1\) and a cladding of index \(n_2\) (\(n_1>n_2\)). Light enters the core from a medium of index \(n_0\) (air, \(n_0=1\)) at angle \(\theta_i\) to the axis, refracts at angle \(\theta_r\), and meets the core–cladding interface at angle \(\phi=90^\circ-\theta_r\).

  n0          cladding n2
 ------------------------------------
   \  θi    /\        /\     core n1
    \      /  \φ    /  \
     \____/____\__/____\____  axis
          θr
 ------------------------------------
              cladding n2

Step 1 (Snell's law at the entry face): \(n_0\sin\theta_i=n_1\sin\theta_r=n_1\sin(90^\circ-\phi)=n_1\cos\phi\)

Step 2 (TIR limit): The ray is just totally reflected when \(\phi=\phi_c\), with \(\sin\phi_c=n_2/n_1\). The corresponding \(\theta_i\) is the maximum, the acceptance angle \(\theta_a\):

\[ n_0\sin\theta_a=n_1\cos\phi_c=n_1\sqrt{1-\sin^2\phi_c}=n_1\sqrt{1-\frac{n_2^2}{n_1^2}}=\sqrt{n_1^2-n_2^2} \]

Step 3: Numerical aperture is defined as \(NA=n_0\sin\theta_a\):

\[ \boxed{NA=\sqrt{n_1^2-n_2^2}} \quad(\text{in air } NA=\sin\theta_a) \]

Using the fractional index change \(\Delta=\dfrac{n_1-n_2}{n_1}\) and \(n_1+n_2\approx2n_1\): \(NA\approx n_1\sqrt{2\Delta}\).

Acceptance angle: \(\theta_a=\sin^{-1}\!\left(\dfrac{\sqrt{n_1^2-n_2^2}}{n_0}\right)\). Rays inside the cone of half-angle \(\theta_a\) (the acceptance cone) propagate by TIR; rays outside it are lost into the cladding.

Physical significance of NA: NA measures the light-gathering ability of the fibre. A larger NA gives a wider acceptance cone and more light coupled in, but also more modes and more intermodal dispersion. NA has no units, and \(0<NA<1\) for a fibre in air (typical values are 0.1 to 0.5).