= Solution
The <Einstein field equations> and the <null energy condition> imply the <null convergence condition>:
$$
R_{ab}U^aU^b=8\pi T_{ab}U^aU^b\geq0.
$$
The scalar-curvature and any <cosmological constant> terms vanish because $U^aU_a=0$. For generators of the <null hypersurface>, the supplied twist-free property gives $\widehat\omega=0$. The <null shear> squared is nonnegative on the screen. Thus the <Null Raychaudhuri equation> implies
$$
\frac{d\theta}{d\lambda}\leq-\frac12\theta^2.
$$
Starting from $\theta(\lambda_0)=\theta_0<0$, the <null expansion> remains negative for as long as the regular congruence exists. Hence
$$
\frac{d}{d\lambda}\left(\frac1\theta\right)
=-\frac{\theta'}{\theta^2}\geq\frac12,\qquad
\frac1{\theta(\lambda)}\geq\frac1{\theta_0}+\frac{\lambda-\lambda_0}{2}.
$$
Before the right-hand side reaches zero, inversion of the negative quantities gives
$$
\theta(\lambda)\leq
\frac{\theta_0}{1+\frac12\theta_0(\lambda-\lambda_0)}.
$$
Since the inverse of a finite negative <null expansion> cannot be nonnegative, the regular congruence cannot continue through the proposed upper limit. The <null focusing theorem> therefore gives
$$
\boxed{\lambda_{\mathrm{focus}}-\lambda_0\leq\frac2{|\theta_0|}.}
$$
Provided the <geodesic> itself extends this far, the transverse area collapses and $\theta\to-\infty$ at or before this bound. An earlier end of the affine <geodesic> would instead be <geodesic incompleteness>. A divergence of the <null expansion> marks a caustic or a <conjugate point to a spacelike surface>; it does not by itself establish a <curvature singularity>.
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