Solution: The equation of the line where the incident light is located is: y=k(x-2) (1).

Let the intersection of the straight line x+y=4 (2) and (1) be E(x, y), then x+k(x-2)=4. x=(4+2k)/(k+ 1)。

Substituting the value of x into (2) can be simplified as: y=2k/k+ 1).

So the coordinates of e are E ((4+2k)/(k+ 1), 2k/(k+ 1)).

Because light is reflected from Q (2,0) through E(x, y) to point F (0,y1) on the Y axis, and then back to point Q (2,0). Therefore, q (2,0) and f (0 0,y1) are two symmetrical points about point e, that is, (2+0)/2=(4+2k)/(k+ 1), which is simplified as: k=-3.

Substitute k=-3 into the coordinate formula of point E to get E( 1, 3).

Because the incident light and the light reflected on the Y axis are symmetrical about a straight line (x+y=4) that passes through E and is vertical. So FQ∨AB.

Let the straight line x+y=4, and the X axis and Y axis intersect at A (4,0) and B(0.4) respectively.

Obtained by similar triangles: OF=2, that is, f (0,2).

The equation of the straight line EF is: x-y+2 = 0-that is, what you want. By using E (1, 3) and F (0, 2), it is simplified to two points.