 Solution

The contrast is weak: $(n_1-n_2)/n_2=0.01$. Use a scalar weak-guidance Helmholtz equation for a transverse field component, with longitudinal wave number $b$ and vacuum wave number $k_0=2\pi /{\lambda }_0$. Separating variables as $\psi =F_l(r)e^{il\varphi }{e}^{ibz}$ givesA bound mode requires $n_2{k}_0<b<n_1{k}_0$. DefineThe step-index fiber normalized frequency is $V$; $u$ is a dimensionless core transverse wave number. Regularity at $r=0$ selects the Bessel function of the first kind, while decay at infinity selects the Modified Bessel function of the second kind:In this scalar approximation, continuity of the field and its radial derivative givesAt cutoff the evanescent wave extends arbitrarily far into the cladding, so $b\downarrow n_2{k}_0$, $w\downarrow 0$ and $u\to V$. For the first azimuthally varying family $l=1$, the recurrence relations reduce the last equality toAs $w\to 0$, $K_1(w)\sim 1/w$ and $K_0(w)\sim -\ln w$, making the right side tend to zero. Thus the onset is at the first positive zero of $J_0$:The scalar $l=0$ fundamental family has no positive cutoff. The first additional spatial family is the $l=1$ family just found; the next radial $l=0$ and first $l=2$ families require the larger zero $j_{1,1}\simeq 3.832$. This establishes the first multimode boundary rather than merely finding one possible cutoff. The two polarizations of the fundamental family are not counted as two spatial modes. Exact vector interface conditions split some higher-mode degeneracies at finite contrast; the displayed condition is the weak-guidance result intended by the supplied scalar equations.