If the equation of mirror is given by `y= 2//pi "sin"pix (y gt 0 , 0 le x le 1`) then find the point on which horizontal ray should be incident so that the reflected ray become perpendicular to the incident ray

Options

A. `((1)/(3) , (sqrt(3))/(pi))`
B. `((sqrt(3))/(pi), (1)/(3))`
C. `((2)/(3) , (sqrt(3))/(pi))`
D. `(1,0)`

Answer

A,C

Explanation

`y = (2)/(pi) "sin"pix, (dy)/(dx)= (2)/(pi) "cos"pixxpiimplies (dy)/(dx) =pm1`
`"tan"45^(@) = 1 = 2 "cos"pix`
`"cos"pix = (1)/(2) implies pix = (npi)/(3) therefore x = (1)/(3) "and" (2)/(3)`
`therefore y = (2)/(pi)xx (sqrt(3))/(2) (because "sin"pi xx (1)/(3)= (sqrt(3))/(2)) = (sqrt(3))/(pi)`