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A beam of light is sent along the line x...

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)`

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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)

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.

A ray of light along x+sqrt(3)y=sqrt(3) gets reflected upon reaching x-axis, the equation of the reflected ray is

The equation of the lines passing through the point (1,0) and at a distance (sqrt(3))/2 from the origin is (a) sqrt(3)x+y-sqrt(3) =0 (b) x+sqrt(3)y-sqrt(3)=0 (c) sqrt(3)x-y-sqrt(3)=0 (d) x-sqrt(3)y-sqrt(3)=0

The director circle of a hyperbola is x^(2) + y^(2) - 4y =0 . One end of the major axis is (2,0) then a focus is (a) (sqrt(3),2-sqrt(3)) (b) (-sqrt(3),2+sqrt(3)) (c) (sqrt(6),2-sqrt(6)) (d) (-sqrt(6),2+sqrt(6))

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The equation of the lines passing through the point (1,0) and at a distance (sqrt(3))/2 from the origin is (a). sqrt(3)+y-sqrt(3) =0 (b). x+sqrt(3)y-sqrt(3)=0 (c). sqrt(3)x-y-sqrt(3)=0 (d). x-sqrt(3)y-sqrt(3)=0

The equation of the line which cuts of intercept 4 on x-ais and makes an angle of 60^(@) with the positive direction of the x-axis is (i) y=sqrt(3)(x+4) (ii) y=-sqrt(3)(x-4) (iii) y=sqrt(3)(x-4) (iv) y=-sqrt(3)(x+4)

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