kindly solve this please.........
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Let MN and MP are two plane mirrors be inclined at an ∠PMN=θ.
A light ray AB parallel to MN is incident on MP and the ray after third reflection on MP at B retraces its path as shown in the diagram. Hence CD is perpendicular to MP.
This means ∠BCN=∠DCM=90∘−θ
AB||MN and BC is the intercept.
So ∠PBA=corresponding ∠PMN=θ
And ∠ABC=180∘−2θ
Since AB||MN and BC is the intercept ,
then ∠ABC+∠BCN=180∘
⇒(180∘−2θ)+(90∘−θ) = 180∘
⇒ 3θ = 90∘
⇒θ = 90∘/3
θ = 30∘
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