If an incident ray directed towards a corner reflector as shown- at what angle- does the reflected ray emerge- nbsp- 22-68-44-46-

The correct option is B 68^∘ From figure, Angle of incidence, ∠ i=90^∘ 22^∘=68^∘ ∠ i=∠ r=68^∘ Since nbsp;N_1⊥ N_2, ∠ r+∠α=90^∘ ⇒∠α=90^∘ 68^∘=22^∘ ∴θ=90 α=68^ ...

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Standard XII

Physics

Mirrors at an angle

If an inciden...

Question

If an incident ray directed towards a corner reflector as shown, at what angle $(θ)$ does the reflected ray emerge?

22^{\circ}

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68^{\circ}

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44^{\circ}

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46^{\circ}

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Solution

The correct option is B $68^{\circ}$

From figure,
Angle of incidence, $∠ i = 90^{\circ} - 22^{\circ} = 68^{\circ}$
$∠ i = ∠ r = 68^{\circ}$
Since $N_{1} ⊥ N_{2}$ ,
$∠ r + ∠ α = 90^{\circ}$
$\Rightarrow ∠ α = 90^{\circ} - 68^{\circ} = 22^{\circ}$
$∴ θ = 90 - α = 68^{\circ}$
Hence, option $(b)$ is the correct answer.

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Mirrors at an angle

Standard XII Physics

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If an incident ray directed towards a corner reflector as shown, at what angle(θ) does the reflected ray emerge?

The correct option is B 68∘ From figure, Angle of incidence, ∠i=90∘−22∘=68∘ ∠i=∠r=68∘ Since N1⊥N2, ∠r+∠α=90∘ ⇒∠α=90∘−68∘=22∘ ∴θ=90−α=68∘ Hence, option (b) is the correct answer.

If an incident ray directed towards a corner reflector as shown, at what angle $(θ)$ does the reflected ray emerge?