= Solution
For an affinely parametrized <null geodesic congruence>, put $B^a{}_b=\nabla_bU^a$. Differentiating along a ray and commuting <covariant derivatives> gives
$$
U^c\nabla_c B^a{}_b=-B^a{}_cB^c{}_b+R^a{}_{dcb}U^dU^c,
$$
using $[\nabla_c,\nabla_b]V^a=R^a{}_{dcb}V^d$. The geodesic acceleration term vanishes. Since $U$ is null and affine, $B^a{}_bU^b=0$ and $U_aB^a{}_b=0$. Its trace is therefore the screen trace $\theta$, and the trace of its square equals that of the <optical tensor> squared. Taking the trace, with $R_{db}=R^a{}_{dab}$, yields
$$
\frac{d\theta}{d\lambda}=-\widehat B_{ab}\widehat B^{ba}-R_{ab}U^aU^b.
$$
Now decompose the <optical tensor> into <null expansion>, <null shear> and <null twist>. The symmetric trace-free part is orthogonal to the trace and to the antisymmetric part, while $\widehat\omega_{ab}\widehat\omega^{ba}=-\widehat\omega_{ab}\widehat\omega^{ab}$. Hence
$$
\widehat B_{ab}\widehat B^{ba}=\frac12\theta^2+\widehat\sigma_{ab}\widehat\sigma^{ab}-\widehat\omega_{ab}\widehat\omega^{ab}.
$$
Substitution proves the <Null Raychaudhuri equation>:
$$
\boxed{\frac{d\theta}{d\lambda}=-\frac12\theta^2-\widehat\sigma_{ab}\widehat\sigma^{ab}+\widehat\omega_{ab}\widehat\omega^{ab}-R_{ab}U^aU^b.}
$$
The squares use the positive-definite screen metric. Thus <null shear> contributes to focusing, whereas nonzero <null twist> has the opposite sign.
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