Physics Question regarding angles of reflection?

Question

Can anyone help me out with some guidance in how to approach this problem?

A ray of light strikes a plane mirror at a 45° angle of incidence. The mirror is then rotated by alpha = 7° into the position shown in red in the drawing, while the incident ray is kept fixed.

http://www.webassign.net/CJ/p25-08alt.gif

(a) Through what angle (theta) does the reflected ray rotate?
°

(b) What is the answer to part (a) if the angle of incidence is 60° instead of 45°?
°

Answer 1

I saw the graphic, so the angle of incidence is 45, the angle of reflection before the mirror rotates is 45 degrees.

Now, when the mirror rotates, the angle will be different : 45º -7º

why ?

Just see the graphic, and you will realize that the angle is the rest of the first incidence angle with the angle that the mirror rotates.

Now, you know by optics, that this new angle will be equal to the new reflecting angle : 38 degrees.

Now, if you wanna find theta, then you first of all, must rest 38 degrees minus 7 degrees, and you will obtain the angle made by the ground, the light blue ground and the red ray.

38 - 7 = 31º

now, we know that the first reflectin angle was 45º, and that angle minus 31 degrees will be theta.

Yes, just wath the graphic, and follow the steps, I am using optics rules, and geometry.

then : 45 - 31 = 14º >>> theta.

b) The process is the same, now using the angle of incidence : 60, and the new angle of reflection 60º.

when the mirror rotates the angle of incidence will be 60 - 7 = 53

This angle of incidence is the same angle of reflection 53 degrees. Now as we did before, you must rest 53 degrees minus 7 degrees, to find the angle between the ground and the red ray.

53 - 7 = 46º

Now, as we know, the first incidence angle that is equal to the first reflecting angle was 60 degrees.

60 - 46 = 14 degrees. theta.

Answer 2

2 trouble-free Snell's regulation utility would provide that it refracts at an perspective of 0, then because it hits air, refracts at perspective that would not exist. So could be a trick ?. perhaps if it refracts at perspective of 0, it would not enter the glass?