Exceptional divisor of the Lefschetz pencil

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I am reading a proof due to Zucker showing that integral Hodge conjecture holds for cubic fourfolds.

Let $X\subset \mathbb{CP}^5$ be a cubic fourfold. We consider a Lefschetz pencil $(Y_t)_{t\in\mathbb P^1}$ of hyperplane sections of $X$, giving rise to morphisms a morphism $\tau: Y\to X$ where $Y:=\{(x,t)\in X\times\mathbb P^1: x\in Y_t\}$. Note that $\tau$ is a blowing up of $X$ along the base-locus (here I note $\Sigma$ the base-locus).

It is said that the exceptional divisor is isomorphic to $\Sigma\times \mathbb P^1$. This is where I didn't understand. Does it hold generally for any projective varieties or just for cubic fourfolds?

p.s. I recall here the definition of Lefschetz pencils just in case. Let $X\subset \mathbb P^N$ be a projective variety. Let $\mathbb P^1\subset \mathbb PH^0(X,\mathcal O_X(1))$ be a straight line which gives a pencil of hyperplanes $(Y_t)_{t\in \mathbb P^1}$. Let $\Sigma:=\cap_{t\in\mathbb P^1}Y_t$ be the base-locus. A pencil of hyperplanes $(Y_t)_{t\in \mathbb P^1}$ is called a Lefschetz pencil if the following two conditions are satisfied:

  • The base-locus $\Sigma$ is smooth of cxdimension $2$ in $X$.
  • Every hyperplane section $Y_t$ has at worse a double ordinary point as singularity.
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If $X \subset \mathbb{P}^N$ is any projective variety and $\Sigma \subset X$ is any complete intersection of two hyperplanes in $X$ then the exceptional divisor of the blowup of $\Sigma$ in $X$ is isomorphic to $\Sigma \times \mathbb{P}^1$.