Question

A beam of light ray sent along the line $x-y=1$ , which after refracting from $x$ axis enters the opposite side by turning through ${30}^{\text{o}}$ away from the normal at the point of incidence with $x$ axis. then which of the following options are correct ?

• A. $Y-$ intercept of refracted ray is $2-\sqrt{3}$
• B. $Y-$ intercept of refracted ray is $\sqrt{3}-2$
• C. perpendicular distance from origin to refracted ray is $\frac{\sqrt{3}+1}{2\sqrt{2}}$ units
• D. perpendicular distance from origin to refracted ray is $\frac{\sqrt{3}-1}{2\sqrt{2}}$ units

Answer 1

Answer: (B), (D)

The correct options are
B $Y-$ intercept of refracted ray is $\sqrt{3}-2$
D perpendicular distance from origin to refracted ray is $\frac{\sqrt{3}-1}{2\sqrt{2}}$ units


Refracted ray passes through $(1,0)$ and have an inclination of $({45}^{\text{o}}-{30}^{\text{o}})={15}^{\text{o}}$

So, required equation is
$y-0=\tan {15}^{\text{o}}(x-1)$

$Y$ intercept = $-\tan {15}^{\text{o}}=\sqrt{3}-2$

Refracted equation is

$y=\frac{\sin {15}^{\text{o}}}{\cos {15}^{\text{o}}}(x-1)\Rightarrow x\sin {15}^{\text{o}}-y\cos {15}^{\text{o}}=\sin {15}^{\text{o}}\Rightarrow x\cos {285}^{\text{o}}+y\sin {285}^{\text{o}}=\sin {15}^{\text{o}}$

which is in normal form $x\cos \alpha +y\sin \alpha =p$

hence, perpendicular distance is $p=\sin {15}^{\text{o}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$