Angle relationships of a blazed diffraction grating

I am trying to understand how a blazed diffraction grating works and came across a deduction I don't understand. I believe that you don't need to know much about optics to answer this question as it is more geometry related.
I have the diagram below of a diffraction grating with all the relevant angles labeled.

The blue zag is a zoom in of the diffraction grating surface. A beam (orange arrow on left) is hitting a face on the surfaces and reflecting (orange arrow on right) relative to normal of the face which is the dotted line between the arrows.
The deduction made is that θi - θr = 2γ and I can't seem to deduce this. I'll be happy to provide any additional information and any help will be appreciated.

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## \theta_i-\gamma=\gamma-\theta_r ## for angle of incidence =angle of reflection (measured from the normal to the blaze). Simple algebra gives ## \theta_i+\theta_r=2 \gamma ##. The equation they have is in error, unless they assign an opposite sign to ## \theta_r ##. ## \\ ## (In some derivations, they do assign an opposite sign to ## \theta_r ##. The equation for constructive interference between adjacent slits or lines on a grating reads ## m \lambda=d (\sin{\theta_i}\pm \sin{\theta_r} ) ##, (this is the condition/location for a primary maximum in the diffraction pattern), where the sign that is used in the ## \pm ## depends upon the sign convention for how ## \theta_r ## is measured. When they put in this opposite sign, this makes ## \theta_i=\theta_r ## for ## m=0 ##, so that the equation reads ## m \lambda=d (\sin{\theta_i}-\sin{\theta_r} )##).

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NihalRi
## \theta_i-\gamma=\gamma-\theta_r ## for angle of incidence =angle of reflection (measured from the normal to the blaze). Simple algebra gives ## \theta_i+\theta_r=2 \gamma ##. The equation they have is in error, unless they assign an opposite sign to ## \theta_r ##. ## \\ ## (In some derivations, they do assign an opposite sign to ## \theta_r ##. The equation for constructive interference between adjacent slits or lines on a grating reads ## m \lambda=d (\sin{\theta_i}\pm \sin{\theta_r} ) ##, (this is the condition/location for a primary maximum in the diffraction pattern), where the sign that is used in the ## \pm ## depends upon the sign convention for how ## \theta_r ## is measured. When they put in this opposite sign, this makes ## \theta_i=\theta_r ## for ## m=0 ##, so that the equation reads ## m \lambda=d (\sin{\theta_i}-\sin{\theta_r} )##).
Thank you, this makes perfect sense now!

## \theta_i-\gamma=\gamma-\theta_r ## for angle of incidence =angle of reflection (measured from the normal to the blaze). Simple algebra gives ## \theta_i+\theta_r=2 \gamma ##. The equation they have is in error, unless they assign an opposite sign to ## \theta_r ##. ## \\ ## (In some derivations, they do assign an opposite sign to ## \theta_r ##. The equation for constructive interference between adjacent slits or lines on a grating reads ## m \lambda=d (\sin{\theta_i}\pm \sin{\theta_r} ) ##, (this is the condition/location for a primary maximum in the diffraction pattern), where the sign that is used in the ## \pm ## depends upon the sign convention for how ## \theta_r ## is measured. When they put in this opposite sign, this makes ## \theta_i=\theta_r ## for ## m=0 ##, so that the equation reads ## m \lambda=d (\sin{\theta_i}-\sin{\theta_r} )##).
On another note, I'm likely to make this another thread but my aim is to uses the grating to combine two laser beams of different wavelengths. I want to do this in an energy efficient manner. Do you know any good resources I could use that explains the math of a blazed grating well?

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On another note, I'm likely to make this another thread but my aim is to uses the grating to combine two laser beams of different wavelengths. I want to do this in an energy efficient manner. Do you know any good resources I could use that explains the math of a blazed grating well?
Your idea sounds like something that could work. Meanwhile I don't have any good source at present that discusses blazed gratings. You would need to google the topic. ## \\ ## For a blazed grating, the peak occurs when incident angle equals angle of reflection off of the blaze, but there is a spread of angles that can get a similar intensity, basically as a result of the single slit diffraction pattern width that occurs, as a multiplying factor of the multi-slit or multi-line interference result from the grating. See also: https://www.physicsforums.com/threa...ratio-of-power-densities.956805/#post-6066788 along with the "link" in post 7 of that thread which I will repeat here: http://www.physics.louisville.edu/cldavis/phys299/notes/lo_msgratings.html

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NihalRi
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@NihalRi Be sure and see the additions that I added to post 5 above.

NihalRi
Your idea sounds like something that could work. Meanwhile I don't have any good source at present that discusses blazed gratings. You would need to google the topic. ## \\ ## For a blazed grating, the peak occurs when incident angle equals angle of reflection off of the blaze, but there is a spread of angles that can get a similar intensity, basically as a result of the single slit diffraction pattern width that occurs, as a multiplying factor of the multi-slit or multi-line interference result from the grating. See also: https://www.physicsforums.com/threa...ratio-of-power-densities.956805/#post-6066788 along with the "link" in post 7 of that thread which I will repeat here: http://www.physics.louisville.edu/cldavis/phys299/notes/lo_msgratings.html
These look really helpful I will have a look into them. I had been reading that the Littrow configuration you mentioned had high efficiency. However wouldn't the order with the most energy be reflected back on the incidence beam and due to the conservation of energy the other orders be less intense?