Answer
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Hint: Draw a diagram showing the value of angles given. As we all know that the angle of incidence will be equivalent to the angle of reflection. Therefore the angle between the reflected ray and the mirror ${{M}_{2}}$ will be the same as the angle between the mirrors. The sum of the angle of a triangle will be equivalent to \[180{}^\circ \]. This will help you in answering this question.
Complete answer:
First of all, draw a diagram showing the value of angles that we know already as per the question.
The angle between the two mirrors are mentioned as,
$\theta =50{}^\circ $
The angle of incidence of the first ray coming has been given as $\alpha $.
As we all know that the angle of incidence will be equivalent to the angle of reflection in general. Therefore the angle of reflection of that ray from the mirror ${{M}_{1}}$will also be $\alpha $.
As this angle is $\alpha $, the angle ABC will be the difference of the $90{}^\circ $ and $\alpha $. That is $90-\alpha $.
As it is already mentioned in the question that the final reflected ray is parallel to the mirror ${{M}_{1}}$. Therefore the angle between the reflected ray and the mirror will be the same as the angle between the mirrors. That is \[50{}^\circ \].
Therefore the angle of reflection of this ray will be equivalent to the difference between $90{}^\circ $ and \[50{}^\circ \]. That is \[40{}^\circ \].
Thus angle of incidence will also be \[40{}^\circ \]. And hence the balance angle will also be \[50{}^\circ \].
As we all know, the sum of the angle of a triangle will be equivalent to \[180{}^\circ \]. Therefore we can write that,
\[\begin{align}
& 180{}^\circ =50{}^\circ +50{}^\circ +90-\alpha \\
& \therefore \alpha =10{}^\circ \\
\end{align}\]
Hence the value of angle of incidence of the first ray has been obtained.
Note:
The angle of incidence can be measured by taking the angle between the incident ray and the normal of the surface. The angle of reflection can be measured by taking the angle between the ray of reflection and the normal of the surface. They are found to be equal in accordance with the laws of reflection.
Complete answer:
First of all, draw a diagram showing the value of angles that we know already as per the question.
The angle between the two mirrors are mentioned as,
$\theta =50{}^\circ $
The angle of incidence of the first ray coming has been given as $\alpha $.
As we all know that the angle of incidence will be equivalent to the angle of reflection in general. Therefore the angle of reflection of that ray from the mirror ${{M}_{1}}$will also be $\alpha $.
As this angle is $\alpha $, the angle ABC will be the difference of the $90{}^\circ $ and $\alpha $. That is $90-\alpha $.
As it is already mentioned in the question that the final reflected ray is parallel to the mirror ${{M}_{1}}$. Therefore the angle between the reflected ray and the mirror will be the same as the angle between the mirrors. That is \[50{}^\circ \].
Therefore the angle of reflection of this ray will be equivalent to the difference between $90{}^\circ $ and \[50{}^\circ \]. That is \[40{}^\circ \].
Thus angle of incidence will also be \[40{}^\circ \]. And hence the balance angle will also be \[50{}^\circ \].
As we all know, the sum of the angle of a triangle will be equivalent to \[180{}^\circ \]. Therefore we can write that,
\[\begin{align}
& 180{}^\circ =50{}^\circ +50{}^\circ +90-\alpha \\
& \therefore \alpha =10{}^\circ \\
\end{align}\]
Hence the value of angle of incidence of the first ray has been obtained.
Note:
The angle of incidence can be measured by taking the angle between the incident ray and the normal of the surface. The angle of reflection can be measured by taking the angle between the ray of reflection and the normal of the surface. They are found to be equal in accordance with the laws of reflection.
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