$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$
We decompose function on $\mathbb{R}^n$ into monochromatic plane waves: \begin{gather*} u(\boldsymbol{x})=\iiint \hat{u}(\xi) e^{i\boldsymbol{x}\cdot \xi}\,d\xi, \end{gather*} where \begin{gather*} \hat{u}(\xi,t)=(2\pi)^{-n}\iiint u(\boldsymbol{x},t) e^{-i\boldsymbol{x}\cdot \xi}\,dx. \end{gather*}
Introducing spherical coordinates in $\mathbb{R}^n\ni \xi$, $\xi=\rho \omega$ with $\rho=|\xi|$ and $\omega\in \mathbb{S}^{n-1}$, which is $(n-1)$-dimensional sphere of radious $1$ in $\mathbb{R}^n$, we see that for odd $n$ \begin{equation} u(\boldsymbol{x})=\frac{1}{2}\int_{-\infty}^\infty \iint_{ \mathbb{S}^{n-1}} \hat{u}(\rho \omega)e^{i\rho \boldsymbol{x}\cdot \omega} \rho ^{n-1}\,d\omega , \end{equation} where we replaced $\int_0^\infty \,d\rho$ by $\frac{1}{2}\int_{-\infty}^\infty \,d\rho$, because $(\rho,\omega)\mapsto (-\rho,-\omega)$ does not change the integrand.
Then \begin{equation} u(\boldsymbol{x})=\iint_{\mathbb{S}^{n-1}} v(\boldsymbol{x}\cdot\omega,\omega)\,d\omega, \label{eq-9.A.2} \end{equation} with \begin{multline} v(s,\omega)=\frac{1}{2}\int_{-\infty}^{\infty} \hat{u}(\rho\omega)e^{i\rho s}\rho^{n-1}\,d\rho=\\ \frac{1}{2}(2\pi)^{-n} \iiint \Bigl(\int \rho^{n-1} e^{-i\rho \boldsymbol{x}\cdot\omega+i\rho s} \,d\rho\Bigr)u(\boldsymbol{x})\,dx=\\ \frac{1}{2}(2\pi)^{1-n}(-i\partial_s)^{n-1} \iiint \delta(x\cdot\omega -s) u(\boldsymbol{x})\,dx=\\ \frac{1}{2}(2\pi)^{1-n}(-i\partial_s)^{n-1} \iint_{\boldsymbol{x}\cdot \omega =s } u(\boldsymbol{x})\,d\Sigma, \label{eq-9.A.3} \end{multline} where $d\Sigma$ is an area element on the $(n-1)$-dimensional plane $\{x\colon x\cdot\omega=s\}$. Here we used that \begin{gather*} \int e^{-i(s'-s)\rho }\,d\rho=2\pi \delta (s-s'). \end{gather*}
Definition \begin{gather} (Ru)(s ; \omega) :=\iint_{\boldsymbol{x}\cdot \omega =s } u(\boldsymbol{x})\,d\Sigma \end{gather} with $s\in \mathbb{R}$ and $\omega\in \mathbb{S}^{n-1}$ is a Radon transform of $u(x)$.
Then \begin{gather} v(s,\omega)= \frac{1}{2}(2\pi)^{1-n}(-i\partial_s)^{n-1} (Ru)(s ; \omega). \label{eq-9.A.4} \end{gather}
Formulae (\ref{eq-9.A.2}) and (\ref{eq-9.A.4}) provide the inverse Radon transform.
Let $u(\boldsymbol{x},t)$ satisfy $n$-dimensional wave equation \begin{gather} u_{tt}-c^2\Delta u =0. \label{eq-9.A.5} \end{gather}
Then its Radon transform $(Ru)(s,\omega,t)$ satisfies $1$-dimensional wave equation \begin{gather} (Ru)_{tt}-c^2(Ru)_ss =0. \label{eq-9.A.6} \end{gather} Therefore, if $u(\boldsymbol{x},0)=0$ \begin{gather} (Ru)(s,\omega,t)= \frac{1}{2c}\int_{s-ct}^{s+ct} (R\phi)(s',\omega)\,ds',\qquad \phi (x)=u_t(x,0), \label{eq-9.A.7} \end{gather} Let $n=3$, define $v(s,\omega)$ by (\ref{eq-9.A.3}): \begin{multline*} v(s,\omega,t) = -\frac{1}{2}(2\pi)^{-2}(\partial_s)^{2} (Ru)(s, \omega,t)= -\frac{1}{16c}\partial_s \Bigl( (R\phi)(s+ct,\omega) - (R\phi)(s-ct,\omega)\Bigr)\\ =-\frac{1}{16c^2\pi^2}\partial_t \Bigl( (R\phi)(s+ct,\omega) + (R\phi)(s-ct,\omega)\Bigr) \end{multline*} and according to (\ref{eq-9.A.2}) \begin{gather*} u(\boldsymbol{x},t)=-\frac{1}{16c^2\pi^2}\partial_t \iint_{\mathbb{S}^2} \Bigl( (R\phi)(\boldsymbol{x}\cdot\omega+ct,\omega) + (R\phi)(\boldsymbol{x}\cdot\omega-ct,\omega)\Bigr)\,d\omega. \end{gather*} From this one can derive (9.1.4).
The CT Scan where (\lambda(s)) is the linear attenuation coefficient (often informally called “optical density”) along the ray. Since the frequency is fixed, (\lambda) is treated as a function of position only.
A larger attenuation (i.e., larger ($\lambda$) produces lower transmitted intensity $I$, which results in a brighter region on a traditional X-ray film (radiographic film), because the film behaves as a negative (reversing darkness). Modern systems use detector plates instead of film, but the mapping from intensity to displayed brightness still follows the same negative-like convention.
Thus, bones—which attenuate X-rays strongly—appear white or light gray, while soft tissues appear dark gray or black.
Note. In classical photographic development, film was produced first and then printed onto photographic paper. Both standard film and paper were negatives: the film reversed the brightness, and the final print reversed it back. *Positive films and papers existed but were later, rarer, and more expensive.
Medical tomography works by “illuminating” the body from many directions perpendicular to a chosen axis (say the $x$-axis) and recording, for each angle, the corresponding Radon transform in the $(y,z)$ plane. By inverting this transform, one reconstructs $\lambda(x; y, z)$, the attenuation coefficient at each point, producing a 3D internal map of the body.
Read more:, Radon transform and CT Scan.