question_answer

A boy \[\mathbf{1}.\mathbf{50}\text{ }\mathbf{m}\] tall with his eye level at \[\mathbf{1}.\mathbf{38}\text{ }\mathbf{m}\] stands before a mirror fixed on a wall. Indicate by means of a ray diagram how the mirror should be positioned so that he can view himself fully. What should be the minimum length of the mirror? Does the answer depend on the eye level?          

Answer:

                The ray diagram is shown in Fig. Here \[AB(=1.5\text{m})\] represents the boy with his head at A, eye level at E and feet at B. Thus \[\text{EB }=\text{ 1}.\text{38 m}\] and \[\text{AE }=\text{ 1}.\text{5}0\text{ }-\text{ 1}.\text{38 }=\text{ }0.\text{12 m}.\] The boy will be able to see his full image. A?B' if the rays starting from head A and feet B reach his eyes after reflection from the mirror MN. Let P and Q be the midpoints of AE and EB, respectively. \[\therefore \] Required minimum length of the mirror is \[MN=PQ=PE+EQ\] \[=\frac{1}{2}AE+\frac{1}{2}EB\] \[=\frac{1}{2}(AE+EB)=\frac{1}{2}AB\] \[=\frac{1.5}{2}\text{m}=\mathbf{0}\mathbf{.75m}\] Thus minimum length of the mirror is always half of the length to be seen and is same for any eye level, but positions of top and bottom edges of the mirror depend on the eye level. For the head to be seen by the eye, the top edge M of the mirror should not be lower than \[PB=\frac{1}{2}AE+EB\] \[=\frac{1}{2}\times 0.12+1.38=1.44\text{m,}\]from the ground. For the foot to be seen, the bottom edge of the mirror should not be higher than \[QB=\frac{1}{2}EB=\frac{1}{2}\times 1.38=0.69m,\]from the ground.