I am interested in finding the minimum width interval $(b_l, b_u)$ for $b_l,b_u \in \mathbb{R}$ that obtains a coverage level of $\alpha \in (0,1)$ for the log-normal distribution. The CDF of the log-normal distribution is given by $\phi(x;\mu,\sigma) = \frac{1}{2}[1 + \text{erf}(\frac{\log(x) - \mu}{\sqrt{2}\sigma})]$. Then any interval satisfying $\phi(b_u) - \phi(b_l) = \alpha$ will contain $(100 \times \alpha)$% of the probability mass.
Therefore, to find the minimum interval width that achieves $\alpha$ coverage I would like to solve:
$$b_l, b_u \text{ that solves } \min_{|b_u - b_l|} \phi(b_u) - \phi(b_l) = \alpha$$
Am I correct in this framing and, if so, what is the solution to this problem? Bonus points for pointers to a reference for the same problem for other probability distributions.