It's not clear which book this is.
For me it was also quite a puzzle to understand the beam splitter until I came to think about it in terms of the most adequate formalism to describe (free) photons, which is QFT and annihilation and creation operators and interpret the beam splitter as some obstacle the photons (aka the electromagnetic waves) get scattered on. Now we assume a lossless beam splitter.
To get a complete description of the beam splitter you have to describe what hapens to a photon in each of the two in-channels, i.e., to which photons they are scattered. For a lossless beam splitterthis you can describe by a socalled Bogoliubov transformation, i.e., you have the annitilation operators for the two in-states ##\hat{a}_{j}##, ##j \in \{1,2 \}## with the bosonic commutation relations
$$[\hat{a}_j,\hat{a}_k^{\dagger}]=\delta_{jk}, \quad [\hat{a}_j,\hat{a}_k]=[\hat{a}_j^{\dagger},\hat{a}_k^{\dagger}].$$
Now the beam-splitter is a device in the realm of linear optics, i.e., the scattering of the photon modes in the in channel leads to photon modes in the out-channel
$$\hat{b}_k =\sum_j U_{jk} \hat{a}_j,$$
and the ##\hat{b}_k## must also fulfill the bosonic commutation relations as the in-channel annihilation and creation operators. Working this out leads to the conclusion that the matrix ##\hat{U}## must be unitary. Since an overall phase doesn't matter, you can assume that ##\hat{U} \in \mathrm{SU}(2)##.
Now we assume that ##\hat{b}_k## annihilates the state which describe the reflected photon of the state which is annihilated by ##\hat{a}_k##. Then our SU(2)-matrix reads
$$\hat{U}=\begin{pmatrix} r & t \\ t' & r'\end{pmatrix}.$$
Now we have the constraints
$$\hat{U}^{-1}=\hat{U}^{\dagger},$$
i.e., that ##\hat{U}## is unitary and our arbitrary choice of phase
$$\mathrm{det} U=1.$$
This implies
$$\hat{U}^{-1}=\frac{1}{\mathrm{det} \hat{U}} \begin{pmatrix} r' & -t \\ -t' & r \end{pmatrix} = \begin{pmatrix} r' & -t \\ -t' & r \end{pmatrix} = \begin{pmatrix} r^* & t^{\prime *} \\ t^* & r^{\prime *}\end{pmatrix}.$$
Thus we have
$$r'=r^*, \quad t'=-t^*.$$
Thus we have
$$\hat{U}=\begin{pmatrix}r & t \\ -t^* & r^* \end{pmatrix}.$$
Now writing
$$r=\rho \exp(\mathrm{i} \varphi), \quad t=\tau \exp(\mathrm{i} \vartheta)$$
with ##\rho,\tau>0## we have
$$\hat{U}=\begin{pmatrix} \rho \exp(\mathrm{i} \varphi) & \tau \exp(\mathrm{i} \vartheta) \\ -\tau \exp(-\mathrm{i} \vartheta) & \rho \exp(-\mathrm{i}\varphi) \end{pmatrix}.$$
From ##\mathrm{det} \hat{U}=1## we get
$$\rho^2+\tau^2=1 \; \Rightarrow \; \rho=\cos \psi, \quad \tau=\cos \psi, \quad \psi \in [0,\pi/2].$$
Thus
$$\hat{U}=\begin{pmatrix} \cos \psi \exp(\mathrm{i} \varphi) & \sin \psi \exp(\mathrm{i} \vartheta) \\ -\sin \psi \exp(-\mathrm{i} \vartheta) & \cos \psi \exp(-\mathrm{i} \varphi) \end{pmatrix}.$$
Of course the parameters depend on the specific realization of the beam splitter.
The meaning of the parameters in the above formalism is obviously as follows: If you have a photon 1 in the incoming channel, it's probability to be reflected is ##|r|^2=\cos^2 \psi## and being transmitted is ##|t'|^2=|t|^2=\sin^2 \psi##. The same probabilities hold for one photon entering in the state 2. You have a 50:50 chance for ##\psi=\pi/4##, i.e., ##\cos^2 \psi=\cos^2 \psi=1/2##.
The exponential factors are phase shifts, depending on the specific type of beam splitter. If you have a symmetric beam splitter, you have
$$r'=r, \quad t'=t.$$
Now above we have seen that
$$r'=r^*, \quad t'=-t^*.$$
This implies that a symmetric beam splitter is given if ##r^*=r##, i.e., ##r \in \mathbb{R}## and ##t=-t^*##, i.e., ##t \in \mathrm{i} \mathbb{R}##. This implies that a symmetric beam splitter has ##\varphi=0## and ##\vartheta =\pm \pi/2##, leading to
$$U=\begin{pmatrix} \cos \psi & \pm \mathrm{i} \sin \psi \\ \pm \mathrm{i} \sin \psi & \cos \psi \end{pmatrix}.$$
Another simple beam splitter is a glass plate with a thin metal coating on one side. Then for the input photon in state 1 the reflection is on the optical denser medium, giving an additional phase shift of ##\pi## relative to the reflected beam of input state 2, which is reflected on the optical less dense medium. This means that, because ##\text{arg} r= \varphi = \text{arg} r' + \pi=\pi-\varphi##, that ##\varphi=\pi/2##.
Then ##\vartheta## is determined by the relative phase shift between the reflected and the transmitted beam for the incoming photon in state 1. It is given by the $$\pi-\vartheta-\varphi=\pi/2-\vartheta =-(n-1) L/\lambda$$, where ##L## is the distance a light beam has to travel in the glass and ##\lambda## the wavelength. This gives ##\vartheta=\pi/2+(n-1)L/\lambda##.
Now let's see, how your book's matrix fits in. It is fine, but has another phase convention since ##\mathrm{det} U_{\text{BS}}=-1,##
i.e., it's related to the ##\hat{U}## in our convention by an additional phase factor ##\exp(\mathrm{i} \pi/2)=\mathrm{i},##
i.e.,
$$\hat{U}=\mathrm{i} \hat{U}_{\text{BS}} = \frac{1}{\sqrt{2}} \begin{pmatrix} \mathrm{i} & \mathrm{i} \\ \mathrm{i} & -\mathrm{i} \end{pmatrix}.$$
This is given by the parameters ##\psi=\pi/4##, ##\varphi=\pi/2##, and ##\vartheta=\pi/2##.