Question

kindly solve this please.........​

Answer 1

Answer:

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∘