Question

If \( \vec{A}, \vec{B} \text{ and } \vec{C} \) are the unit vectors along the incident ray, reflected ray and outward normal to the reflecting surface, then:

• A. \( \vec{B} = \vec{A} - \vec{C} \)
• B. \( \vec{B} = \vec{A} + (\vec{A}\cdot\vec{C})\vec{C} \)
• C. \( \vec{B} = \vec{A} + \vec{C} \)
• D. \( \vec{B} = \vec{A} - 2(\vec{A}\cdot\vec{C})\vec{C} \)

Hint:

Reflection flips the component along normal while keeping parallel component unchanged.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Laws of reflection state that the angle of incidence equals the angle of reflection and all rays lie in the same plane. In vector form, only the component of the velocity/ray parallel to the normal is reversed.
: Key Formula or Approach:
The reflected unit vector \( \hat{r} \) in terms of incident vector \( \hat{i} \) and normal vector \( \hat{n} \) is:
\[ \hat{r} = \hat{i} - 2(\hat{i} \cdot \hat{n})\hat{n} \]
Step 2: Detailed Explanation:
- Let \( \vec{A} \) be the incident unit vector.
- Let \( \vec{C} \) be the outward normal unit vector.
- Let \( \vec{B} \) be the reflected unit vector.
The component of \( \vec{A} \) along the normal \( \vec{C} \) is \( (\vec{A} \cdot \vec{C})\vec{C} \).
In a reflection, the component parallel to the surface remains unchanged, but the component perpendicular to the surface (parallel to the normal) is reversed.
Mathematically, we subtract twice the normal component from the original vector to achieve this reversal:
\[ \vec{B} = \vec{A} - 2(\vec{A} \cdot \vec{C})\vec{C} \]
Step 3: Final Answer:
The reflected ray vector is \( \vec{B} = \vec{A} - 2(\vec{A} \cdot \vec{C})\vec{C} \).