$(a)\;30^{\circ} \\ (b)\;60^{\circ} \\ (c)\;180^{\circ} \\ (d)\;90^{\circ} $

Let $\theta$ be the angle between two mirrors $OM_1$ and $OM_2$.

The incident ray $AB$ is parallel to mirror $OM_2$ and strikes the mirror $OM_1$ at an angle of incidence equal box .

It is reflected along BC the angle of reflection is $\alpha$

From figure we have $\angle M_1BA=\angle OBC = \angle M_1OM_2=\theta$

Similarly for reflection at mirror $OM_2$, we have $\angle M_2 CD=\angle BCO=\angle M_2 OM_1 =\theta$

Now in triangle $OBC$

$3 \theta=180$

$\therefore \theta= 60^{\circ}$

Hence b is the correct answer.

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