Question

A ray of light travelling along the line $x+\sqrt{3}y=5$ is incident on the x-axis and after refraction it enters the other side of the x-axis by turning $\frac{\pi}{6}$ away from the x-axis. The equation of the line along which the refracted ray travels is

• A. x+\sqrt{3}y-5\sqrt{3}=0
• B. x-\sqrt{3}y-5\sqrt{3}=0
• C. \sqrt{3}x+y-5\sqrt{3}=0
• D. \sqrt{3}x-y-5\sqrt{3}=0

Answer 1

Answer: (C)

\sqrt{3}x+y-5\sqrt{3}=0

Answer 2

The incident ray is given by $x+\sqrt{3}y=5$. Its slope is $m_i = -\frac{1}{\sqrt{3}}$, which corresponds to an angle $\theta_i = 150^\circ$ with the positive x-axis. The ray is incident on the x-axis ($y=0$), so the point of incidence is $(5, 0)$. The ray turns $\frac{\pi}{6}$ ($30^\circ$) away from the x-axis. This means the angle of deviation is $30^\circ$. The angle of the refracted ray with the positive x-axis, $\theta_r$, can be $\theta_i - 30^\circ$ or $\theta_i + 30^\circ$. $\theta_r = 150^\circ - 30^\circ = 120^\circ$ or $\theta_r = 150^\circ + 30^\circ = 180^\circ$. If $\theta_r = 180^\circ$, the refracted ray is along the x-axis, which is not a refraction into the "other side". Thus, $\theta_r = 120^\circ$. The slope of the refracted ray is $m_r = \tan(120^\circ) = -\sqrt{3}$. The equation of the refracted ray passing through $(5, 0)$ with slope $-\sqrt{3}$ is $y - 0 = -\sqrt{3}(x - 5)$, which simplifies to $y = -\sqrt{3}x + 5\sqrt{3}$, or $\sqrt{3}x + y - 5\sqrt{3} = 0$.