Light rays and interference fringes

[pewpew95]

There is a non-AAMC practice question that goes as follows:

"If a light ray strikes an interference fringe at a 35 degree angle with respect to the normal plane, the reflected angle with respect to the interference fringe will be..." I made an educated guess of the angle being greater than 35 degrees, and got it right.

I'm not quite sure how to correctly solve such a problem. The explanation states the reflected angle is the same as the incident angle but is also complementary to the incident angle, therefore making the reflected angle 55 degrees. I was actually assuming the reflected angle would be 180-35 = 145 degrees. Any help?

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[theonlytycrane]

The interference fringe is just like a flat surface and the 35 degree angle refers to the angle between the incident ray and the normal. The reflected angle with respect to the normal will also be 35 degrees.

The angle they want you to compare the 35 degree angle to is the angle between the reflected ray and the flat surface which would be 90 - 35 = 55.

 

[sh.h]

Is there a figure