Straight Lines
Reflection of a line
Grade 11
Question:
<p>A ray of light is incident on a mirror. The normal at the point of incidence makes an angle of 30° with the horizontal. If the point of incidence is \((\sqrt{3}, 0)\) and a point (0, 1) lies on the ray of light (on the line), then the equation of the reflected ray is:</p>
<p>\(\sqrt{3}y = x + \sqrt{3}\)</p>
<p>\(\sqrt{3}y = x - \sqrt{3}\)</p>
<p>\(y = \sqrt{3}x - \sqrt{3}\)</p>
<p>\(y\sqrt{3} = x + \sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: The reflected ray is obtained by reflecting the incident ray's direction across the normal. Use the angle the normal makes with the horizontal to find the slope of the normal, then apply the reflection formula: if incident ray has slope m₁ and normal has slope m_n, the reflected ray has slope m₂ where the normal bisects the angle between them.
<p><strong>Step 1:</strong> Find the slope of the normal. The normal makes 30° with horizontal, so slope of normal = tan(30°) = 1/√3.</p><p><strong>Step 2:</strong> Find the slope of the incident ray. The incident ray passes through (√3, 0) and (0, 1).</p><p>Slope m₁ = (1 - 0)/(0 - √3) = -1/√3</p><p><strong>Step 3:</strong> Use the reflection formula. If m_n is the normal's slope and m₁ is the incident ray's slope, the reflected ray's slope m₂ is found using:</p><p>m₂ = (2m_n(1 + m₁m_n) - m₁(1 - m_n²))/(1 - m_n² + 2m_nm₁)</p><p>Alternatively, using angle approach: Normal makes 30° with horizontal. Incident ray has slope -1/√3 (makes -30° with horizontal). By law of reflection, the reflected ray makes 90° with incident ray direction about the normal.</p><p><strong>Step 4:</strong> The incident ray makes angle -30° with horizontal. Normal makes 30°. The angle between normal and incident ray is 60°. The reflected ray makes 30° + 60° = 90° - 30° = 60° on the other side, giving slope = tan(60°) = √3.</p><p><strong>Step 5:</strong> Equation of reflected ray through (√3, 0) with slope √3:</p><p>y - 0 = √3(x - √3)</p><p>y = √3x - 3</p><p>∴ Answer: B</p>
Correct Answer: B