Straight Lines
Straight Line
Allen Star Batch
Grade 11
Question:
A ray of light is sent along the line $x - 2y = 8$. After refracting across the line $x + y = 1$ it enters the opposite side after turning by $15°$ away from the line $x + y = 1$. Then the equation of line along which refracted ray travels will:
have slope $\frac{5\sqrt{3} - 6}{3}$
have slope $\frac{5\sqrt{3} - 6}{13}$
pass through $\left(0, \frac{13\sqrt{3} - 50}{3\sqrt{3}}\right)$
pass through $\left(0, \frac{-50\sqrt{3} - 31}{39}\right)$
Step-by-Step Solution
Key Concept: The law of reflection requires equal angles of incidence and reflection, solved using the tangent angle formula between two lines.
From $x - 2y = 8$ and $x + y = 1$, solving simultaneously gives $y = - rac{7}{3}$ and $x = rac{10}{3}$, so $A = ( rac{10}{3}, - rac{7}{3})$. For the refracted ray with slope $m$, the angle of incidence equals angle of reflection. Using $ an 15° = rac{|m - (-1)|}{|1 + m(-1)|} = rac{|2m - 1|}{|m + 2|}$, we get $2m - 1 = (2 - \sqrt{3})(m + 2)$, yielding $m = rac{5\sqrt{3} - 6}{3}$. The angle $\phi$ between $x + y = 1$ and the refracted ray through $A$ with slope $ rac{5\sqrt{3} - 6}{3}$ is found using the angle formula.
Correct Answer: 1,3