\n– The incident electromagnetic wave is given by $\\vec{E} = \\vec{E}_0 e^{i k_0(3x+4y)-\\omega t}$. The wave vector $\\vec{k}_i$ is the vector part multiplying $\\vec{r}=(x,y,z)$ in the exponent (ignoring $k_0$). So, the incident wave vector in air is proportional to $3\\hat{i} + 4\\hat{j}$. We can write $\\vec{k}_i = k_0 (3\\hat{i} + 4\\hat{j})$.
\n– The magnitude of the incident wave vector is $|\\vec{k}_i| = k_0 \\sqrt{3^2 + 4^2} = 5 k_0$. In air, the refractive index $n_1 = 1$, so $|\\vec{k}_i| = n_1 (\\omega\/c) = \\omega\/c$. This implies $k_0 = \\omega\/(5c)$.
\n– The wave is incident on the x-z plane, which is the plane where $y=0$. The normal to this plane is along the $\\hat{j}$ direction. Let’s assume the glass slab is in the region $y>0$, and the air is in $y<0$. The incident wave has $k_{iy} = 4k_0 > 0$, so it is travelling towards the interface $y=0$ from the region $y<0$. The normal vector pointing into the glass is $\\hat{j}$.\n- According to Snell's law for wave vectors, the component of the wave vector parallel to the interface is conserved. The interface is the x-z plane, so the components parallel to the plane are the x and z components.\n- Incident wave vector $\\vec{k}_i = 3k_0 \\hat{i} + 4k_0 \\hat{j} + 0\\hat{k}$. The parallel component is $k_{ix} \\hat{i} + k_{iz} \\hat{k} = 3k_0 \\hat{i}$.\n- The transmitted wave vector in glass is $\\vec{k}_t = k_{tx} \\hat{i} + k_{ty} \\hat{j} + k_{tz} \\hat{k}$.\n- Conservation of parallel component: $k_{tx} = k_{ix} = 3k_0$ and $k_{tz} = k_{iz} = 0$. So $\\vec{k}_t = 3k_0 \\hat{i} + k_{ty} \\hat{j}$.\n- The magnitude of the transmitted wave vector is $|\\vec{k}_t| = n_2 (\\omega\/c) = n_2 n_1^{-1} |\\vec{k}_i| = 1.5 \\times 1^{-1} \\times 5k_0 = 7.5 k_0$.\n- $|\\vec{k}_t|^2 = k_{tx}^2 + k_{ty}^2 = (3k_0)^2 + k_{ty}^2 = (7.5 k_0)^2$.\n- $9k_0^2 + k_{ty}^2 = 56.25 k_0^2$.\n- $k_{ty}^2 = (56.25 - 9) k_0^2 = 47.25 k_0^2 = \\frac{189}{4} k_0^2$.\n- $k_{ty} = \\pm \\sqrt{\\frac{189}{4}} k_0 = \\pm \\frac{3\\sqrt{21}}{2} k_0$.\n- Since the wave is transmitted from air ($y<0$) into glass ($y>0$), the y-component of the transmitted wave vector must be positive (pointing into the glass region). So $k_{ty} = \\frac{3\\sqrt{21}}{2} k_0$.
\n– The transmitted wave vector is $\\vec{k}_t = 3k_0 \\hat{i} + \\frac{3\\sqrt{21}}{2} k_0 \\hat{j}$. The direction of propagation is given by the unit vector in the direction of $\\vec{k}_t$.
\n– The direction is proportional to $3\\hat{i} + \\frac{3\\sqrt{21}}{2} \\hat{j}$.
\n– Let’s check the options. Option A is $0.4\\hat{i} + 0.2\\sqrt{21}\\hat{j} = \\frac{2}{5}\\hat{i} + \\frac{\\sqrt{21}}{5}\\hat{j}$.
\n– Check if $(3, \\frac{3\\sqrt{21}}{2})$ is proportional to $(\\frac{2}{5}, \\frac{\\sqrt{21}}{5})$. The ratio of x-components is $3 \/ (2\/5) = 15\/2$. The ratio of y-components is $(\\frac{3\\sqrt{21}}{2}) \/ (\\frac{\\sqrt{21}}{5}) = \\frac{3\\sqrt{21}}{2} \\times \\frac{5}{\\sqrt{21}} = \\frac{15}{2}$.
\n– Since the ratios are equal, the vectors are proportional. The direction of propagation of the transmitted wave is proportional to $0.4\\hat{i} + 0.2\\sqrt{21}\\hat{j}$.
\n<\/section>\n