1. Key idea: stratified medium
   The refractive index inside the slab varies only with depth $z$, i.e. $n=n(z)$.
   Such a medium is layered parallel to the slab faces, so at every horizontal layer the quantity $n(z)\,\sin\theta(z)$ remains constant along the ray.

2. Apply Snell's law across all horizontal layers
   At the top interface (medium 1 to slab), $n_1\sin\theta_i = n(0)\sin\theta(0).$ Inside the slab, since the medium changes only with $z$, $n(z)\sin\theta(z)=\text{constant}.$ At the bottom interface (slab to medium 2), this same constant must equal $n_2\sin\theta_f.$ Therefore, $n_1\sin\theta_i = n_2\sin\theta_f.$ Hence option C is true.

   Option B says $n_1\sin\theta_i=(n_2-n_1)\sin\theta_f,$ which is not the correct refraction relation. So B is false.

3. Expression for lateral shift
   Let the ray make angle $\theta(z)$ with the normal inside the slab. Then $\sin\theta(z)=\frac{n_1\sin\theta_i}{n(z)}$ using the constant obtained above.

   The horizontal displacement while crossing a thickness element $dz$ is $dx=\tan\theta(z)\,dz.$ So total lateral displacement is $l=\int_0^d \tan\theta(z)\,dz.$ Since $\theta(z)$ depends on $n(z)$, we get $l=\int_0^d \tan\left(\sin^{-1}\frac{n_1\sin\theta_i}{n(z)}\right)dz,$ so clearly $l$ depends on the profile $n(z)$. Hence option A is true.

4. Does $l$ depend on $n_2$?
   Inside the slab, the path is completely determined by the invariant $n(z)\sin\theta(z)=n_1\sin\theta_i,$ which depends on the incident conditions and the slab profile $n(z)$, but not on $n_2$.

   The medium $n_2$ only affects the final emergence angle through $n_2\sin\theta_f=n_1\sin\theta_i,$ but the accumulated horizontal shift inside the slab, $l=\int_0^d \tan\theta(z)\,dz,$ is already fixed before emergence. Therefore $l$ is independent of $n_2$, so option D is true.

5. Final evaluation of options

   • A: True
   • B: False
   • C: True
   • D: True

Therefore, the correct statements are $\boxed{A,\ C,\ D}.$