A paralel beam is incident on a prism.the prism has vertical angle 4`and base angles 88`&88` respectively. A Plane miror is placed verticaly on the other side of incoming rays. By what angle shld the miror be rotated sop that light returns back to its original path.(refractive index of prism is 1.5) {ans.....2 degree}

draw a prism ABC with verticle angle A =4 degree...

now draw a plane mirror which is inclined at @ with horizontal plane...

the ray returns to its original path if & if only when the ray coming out from the prism is normal to the mirror...

now draw this diagram

we have d(angle of deviation) = (n-1)A                              (formula)

                                           =2^o                         ............1       (n=3/2)

let ray intersects the prism at X & comes out , & intersects plane mirror at Y and is reflected back ,

now  extend the parallel ray which was incident on prism, so that it touches the plane mirror at point Z...

now we have XYZ as a right angled triangle....

angle X = angle of deviation

angle Z = @                             (from diagram)

angle Y =90

X+Y+Z =180

d + @ + 90 = 180

d+@ = 90

@ = 88^o                   ...................2

initially mirror was at 90^o , but when its angle is 88^o our purpose is solved so it should be rotated through (90-88)=2^o