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Four indentical mirrors are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint M of mirror AD and after two reflections reaches corner D. Then, angle theta must be |
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Answer» `tan^(-1)(0.75)` From figure,  `tantheta=(x)/((a//2))` ...(i) `tantheta=(a-x)/(y)` ...(ii) `tantheta=(a)/(a-y)` ...(iii) From (i) and (ii), we get `(2xy)/(a)=(a-x)/(y),2xy=a^(2)-xa` ...(iv) From (ii) and (iii), we get `(a-x)/(y)=(a)/(a-y)impliesa^(2)-ya-xa+xy=ya` `a^(2)-xa-ya+xy=ya` `3xy=2ayimpliesx=(2a)/(3)` (Using(iv)) Substituting this value of x in equation (i), we get `tantheta=((2a//3))/((a//2))=(4)/(3) :. cottheta=(1)/(tantheta)=(3)/(4)` or `theta=cot^(-1)(0.75)` |
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