Question

A linearly polarized light beam travels from origin to point A (1,0,0). At the point A, the light is reflected by a mirror towards point B (1, -1,0). A second mirror located at point B then reflects the light towards point C (1,-1,1). Let n(x, y, z) represent the direction of polarization of light at (x, y, z).

• A. If \(\hat{n}\)(0, 0, 0) = \(\hat{y}\), then \(\hat{n}\)(1, -1, 1) = \(\hat{x}\)
• B. If \(\hat{n}\)(0, 0, 0) = \(\hat{z}\), then \(\hat{n}\)(1, -1, 1) = \(\hat{y}\)
• C. If \(\hat{n}\)(0, 0, 0) = \(\hat{y}\), then \(\hat{n}\)(1, -1, 1) = \(\hat{y}\)
• D. If \(\hat{n}\)(0, 0, 0) = \(\hat{z}\), then \(\hat{n}\)(1, -1, 1) = \(\hat{x}\)

Answer 1

The Correct Option is A, B

The problem involves analyzing how the polarization direction of light changes as it reflects off mirrors positioned at specific points in a coordinate system.

Step 1: Understand the reflection condition at point A.

The light is originally traveling along the \( x \)-axis from the origin to point A (1, 0, 0). At point A, it is reflected towards point B (1, -1, 0). This means the light path changes from moving in the positive \( y \)-direction to moving in the negative \( y \)-direction upon reflection. According to the laws of reflection, the plane of incidence is the \( xy \)-plane.

Step 2: Analyze the polarization change at point A.

If the polarization direction \( \hat{n}(0, 0, 0) = \hat{y} \), the polarization direction remains unchanged after reflection because it lies in the plane of incidence. However, if \( \hat{n}(0, 0, 0) = \hat{z} \), the polarization direction stays unchanged as it is perpendicular to the plane of incidence.

Step 3: Consider the second reflection at point B.

After reaching point B, the light reflects towards point C (1, -1, 1). Here, the plane of incidence includes the \( yz \)-plane as the light now moves from \( (1, -1, 0) \) to \( (1, -1, 1) \) along the \( z \)-axis.

For the initial condition \( \hat{n}(0, 0, 0) = \hat{y} \), the direction of polarization does not change since \( \hat{y} \) remains in the \( yz \)-plane, the new plane of incidence.

For the initial condition \( \hat{n}(0, 0, 0) = \hat{z} \), the \( \hat{z} \) component lies along the incident direction and does not contribute to the polarization after reflection, thus resulting in a switch to the \( \hat{y} \) direction.

Conclusion:

• If \(\hat{n}(0, 0, 0) = \hat{y}\), the polarization direction at (1, -1, 1) becomes \(\hat{x}\).
• If \(\hat{n}(0, 0, 0) = \hat{z}\), the polarization direction at (1, -1, 1) becomes \(\hat{y}\).