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Question
Prove geometrically that when plane mirror turns through a certain angle, the reflected ray turns through twice the angle.
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Solution
Consider a ray of light AB, incident on a plane mirror in position MM’, such that BC is the reflected ray and BN is normal.
∠ABM = ∠CBN = ∠i
∠ABC = 2 ∠i …(i)
Let the mirror be rotated through an angle ‘0’ about point B, such that M1M1 is the new position of the mirror and BN1 is the new position of normal. As the position of the incident ray remains same, therefore new angle of the incidence is ∠ABN1 whose magnitude is (i + θ). Let BD be the reflected ray, such that ∠DNB1 is the new angle of reflection.
∠ABD = ∠ABN1 + ∠DBN1
= ∠(i + θ) + ∠(i + θ)
= 2 ∠i + 2∠θ
Subtracting (i) from (ii)
∠ABD - ∠ABC = 2∠i + 2∠θ - 2∠i
∴ ∠CBD = 2∠θ
Thus, for a given incident ray, if the plane mirror is rotated through a certain angle, then the reflected ray rotates through twice the angle.
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