F4 Phsyic Past Paper !(2)

First draw the image of the pole P'Q' , which is at a distance of (4 + 1) m = 5 m behind the mirror.

Now join P'and E, which should passes through M in order for the eye at E to see the top of the pole.
Similarly, join Q' to E, which should passes through N in order for the eye at E to see the bottom of the pole.

Hence, triangle EMN and triangle EP'Q' are similar triangles. Since the eye E is 1 m from the mirror MN and (5+1) m = 6 m from the image of the pole P'Q'
MN/1 = P'Q'/6
h/1 = 3/6
i.e. h = 0.5 m

圖: http://s16.postimage.org/4nsnway9f/physics.png

AF = image of the wall
DG = the mirror

Consider Triangle EDG and EAF (which are similar triangle),
DG:AF = CB:CA
>> h:3 = 1:6

Therefore, h = 0.5m