Physics Q45 – JEE-MAIN 2026

An unpolarized light is incident on the plane interface of air-dielectric medium shown in figure. If the incident angle is equal to Brewster angle, identify the expression representing reflected wave.

Clue 1 of 3: Brewster's Angle Polarization

At Brewster's angle, the reflected light is completely linearly polarized.

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Step 1: Identify Plane of Incidence and Reflected Polarization✦ Active

From the diagram, the incident light is shown in the x-z plane. Therefore, the plane of incidence is the x-z plane. At Brewster's angle, the reflected light is linearly polarized perpendicular to the plane of incidence. This means the electric field vector of the reflected wave must be along the y-axis ( $\hat{j}$ ). Thus, the vector part of the electric field expression should be $E_{y} \hat{j}$ (implying that the x and z components of the reflected electric field, $E_{x}$ and $E_{z}$ , are zero).

Step 2: Determine Reflected Wave Vector Direction○ Expand

The interface is the x-y plane ( $z = 0$ ). The z-axis points downwards into the medium. The incident wave comes from air (above the interface) and moves towards $z = 0$ . The reflected wave moves away from the interface, meaning it propagates in the negative z-direction. Since the incidence is in the x-z plane, the reflected wave vector will have components in the x and z directions. So, the reflected wave vector is of the form ${\vec{k}}_{r} = k_{x} \hat{i} - k_{z} \hat{k}$ (where $k_{x}$ and $k_{z}$ are positive magnitudes). The phase term for such a wave is ${\vec{k}}_{r} \cdot \vec{r} - ω t = (k_{x} x - k_{z} z - ω t)$ .

Step 3: Match with Options○ Expand

Based on the polarization, we look for an option that reduces to $E_{y} \hat{j}$ when $E_{x} = 0$ and $E_{z} = 0$ . Options 1 and 4 partially fit this. Let's analyze them further:

Option 1: $(E_{x} \hat{i} + E_{y} \hat{j}) \sin (k x - k z - ω t)$ . If $E_{x} = 0$ (for the reflected wave), the vector part is $E_{y} \hat{j}$ . The phase term $(k x - k z - ω t)$ is consistent with the form $(k_{x} x - k_{z} z - ω t)$ for the reflected wave.

Option 4: $(E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k}) \sin (k x + k y - k z - ω t)$ . While the vector part can reduce to $E_{y} \hat{j}$ (if $E_{x} = 0$ and $E_{z} = 0$ ), the phase term $(k x + k y - k z - ω t)$ includes a $k y$ component. This implies propagation in the y-direction, which is incorrect for reflection in the x-z plane.

Therefore, option 1 is the most consistent expression representing the reflected wave at Brewster's angle.

💡 Teacher's Secret Hint

Remember that $E_{x}, E_{y}, E_{z}$ in the options represent the components of the electric field of the *reflected* wave, which are determined by the polarization state.

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Physics Question 45 – JEE-MAIN 2026

Clue 1 of 3: Brewster's Angle Polarization

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