A ray of light is incident on a pane mirror at a certain point. The mirror is capable of rotation about an axis passing through the same point. Prove that if the mirror turns through a certain angle then the reflected ray turns through double the angle.
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Let, XY be the position of mirror initially. A ray AO reflected at O goes along OB.
∠ AON = ∠ BON
∴ ∠ AOY =∠ BOX = α (∵assume)
Let, the mirror be rotated to a new position XY' through an angle ∅ about an axis passing through X. The incident ray, now, gets reflected from O' along O'B'. O'N' is the new position of normal.
According to the law of reflection,
∠AO'N = ∠ B'O'N'
∴ ∠ AO'Y' = ∠B'O'X' = α
Also, ∠ XO'O = ∠ AO'Y' = α
∴ ∠B'O'O = 2α'
Due to rotation of mirror the reflected ray turns through an angle BCB' = θ.
∠ BCB' = ∠ O'CO = θ
∠ O'OC = 180° - 2α
In Δ OO'C, θ + 180° - 2α + 2α' = 180°
or, θ = 2(α-α')
In Δ XOO' , α = ∅ + α'
∴ ∅ = α - α'
Substituting for ∅ in (i), we get θ = 2∅
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