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Two plane mirrors are inclined to each other in such a way that a ray of light moving parallel to one mirror comes out parallel to the other mirror after two reflections. The angle between the mirrors is :

$(a)\;30^{\circ} \\ (b)\;45^{\circ} \\ (c)\;90^{\circ} \\ (d)\;60^{\circ} $

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Since $N_1$ is normal to mirror $M_1$
angle of incidence = angle of reflection
and $AB$ || on mirror $M_2$
$\angle ABM_1 =\angle CBO= \theta$
Similarly $N_2$ is normal to mirror $M_2$
$\angle M_2 CE =\angle CB0 =\theta$
In $\Delta OCB; $
$3 \theta= 180^{\circ}$
$\theta =60^{\circ}$
Hence d is the correct answer.
answered Jan 15, 2014 by meena.p

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