Show that if a ray of light is reflected successively from two mirrors inclined at an angle $θ$ , the deviation of the ray does not depend upon the angle of incidence. [5 MARKS]

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Solution

Diagram: 2 Marks
Problem: 3 Marks

If $d_{1} a n d d_{2}$ are the deviations suffered at $M_{1} a n d M_{2}$ respectively.
Then $d_{1} =$ $π - 2i$ and $d_{2} =$ $π - 2i'$

Total deviation $d = d_{1} + d_{2}$ , $d = 2π - 2(i + i`) . . . . . . . (1)$

From triangle OAB, $θ +$ $\frac{π}{2} - i + \frac{π}{2} - i^{'} = π$ [ $∵$ sum of all angles of triangle is $180^{o}$ ( $π$ )]
$θ = i + i^{'} + π - π$
So, $θ = i + i^{'}$
Putting value in eq(1)
Therefore, total deviation, $d = 2π - 2θ$

Thus the total deviation is independent of angle of incidence.

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Show that if a ray of light is reflected successively from two mirrors inclined at an angle θ , the deviation of the ray does not depend upon the angle of incidence. [5 MARKS]

Show that if a ray of light is reflected successively from two mirrors inclined at an angle $θ$ , the deviation of the ray does not depend upon the angle of incidence. [5 MARKS]