A ray of light is sent along the line `2x-3y=5.` After refracting across the line `x+y=1` it enters the opposite side after torning by `15^0` away from the line `x+y=1` . Find the equation of the line along which the refracted ray travels.

If a ray travelling the line x = 1 gets reflected the line x+y=1 then the equation of the line along which the reflected ray travels is

A ray of light travelling along the line x+y=1 is incident on the X - axis and after refraction the other side of the X - axis by turning pi//6 by turning away from the X - axis .The equation of the line along which the refracted ray travels is

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 (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)

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)

If a ray travelling along the line x = 1 gets reflected from the line x + y = 1, then the eqaution the line along which the reflected ray travels is

A ray of light is sent along the line x-2y-3=0 upon reaching the line 3x-2y-5=0, the ray is reflected from it. Find the equation of the line containing the reflected ray.

A ray of light is sent along the line x-2y-3=0 upon reaching the line 3x-2y-5=0, the ray is reflected from it. Find the equation of the line containing the reflected ray.

A ray of light is sent along the line x-2y-3=0 upon reaching the line 3x-2y-5=0, the ray is reflected from it. Find the equation of the line containing the reflected ray.

A ray of light is sent along the line x-2y+5 = 0 upon reaching the line 3x- 2y + 7 = 0 the ray is reflected from it . Find the equation of the containing the reflected ray .

A ray of light is incident along a line which meets another line, 7x-y+1 = 0 , at the point (0, 1). The ray isthen reflected from this point along the line, y+ 2x=1 . Then the equation of the line of incidence of the ray of light is (A) 41x + 38 y - 38 =0 (B) 41 x - 38 y + 38 = 0 (C) 41x + 25 y - 25 = 0 (D) 41x - 25y + 25 =0