Correct Answer - A
Here, `vec A = (1)/(2)(hat i+ sqrt(3) hat j)`
and `vec B = (1)/(2)(hat i- sqrt(3) hat j)`
`:. A=B=sqrt(I^2+(sqrt(3))^2)=2` and `vec A. vec B = -2`
`cos theta = (vec A. vec B)/(A B) = (-2)/(2 xx 2) = - (1)/(2)`
`theta = 120^@`, Fig.
Angle of incidence,
`i = (180^@ - theta)/(2) = (180^@ - 120^@)/(2) = 30^@`.
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