A beam of light is sent along the line `x-y=1` , which after refracting from the x-axis enters the opposite side by turning through `30^0` towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is `(2-sqrt(3))x-y=2+sqrt(3)` `(2+sqrt(3))x-y=2+sqrt(3)` `(2-sqrt(3))x+y=(2+sqrt(3))` `y=(2-sqrt(3))(x-1)`

Options

A. `(2-sqrt(3))x-y=2+sqrt(3)`
B. `(2+sqrt(3))x-y=2+sqrt(3)`
C. `(2-sqrt(3))x+y=(2+sqrt(3))`
D. `y=(2+sqrt(3))(x-1)`

Answer

D

Explanation

From the figure, the refracted ray makes and angle of `15^(@)` with the positive direction of the x-axis and passes through the point (1,0). Its equation is
`(y-0)=("tan" 15^(@))(x-1)`
`"or "y=(2-sqrt(3))(x-1)`