Question

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.

[no copy or spam or improper is answer is allowed!!
any kind of violation will be reported so be careful, if you don't know then don't reply just quite]

Answer 1

Answer:

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∅

Answer 2

Answer:

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∅