Angle of Deflection of Laser Given Tilted Mirror

AI Thread Summary
The discussion centers on proving that a laser beam deflected by a rotated mirror changes its angle by 2θ, where θ is the angle of rotation. The initial confusion arises from the relationship between the angles of incidence and reflection, with participants clarifying that the angle of deflection (γ) is indeed equal to twice the angle of rotation (2θ). Through geometric reasoning, it is established that when the mirror is tilted, the relationship holds true as the reflected angle combines both the incident and deflected angles. The conclusion confirms that the angle of deflection is correctly represented as 2θ, resolving the initial doubts. This discussion highlights the importance of understanding geometric relationships in optics.
Kumo
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Hi Everyone,

A question in my latest Physics I lab wishes for a proof showing that if a laser beam is incident on a mirror that is then rotated an angle θ the beam is deflected an angle 2θ.

I attempted to prove this geometrically below. In the diagram the angle γ, is the angle of deflection, and β is the incident angle. I was able to show that if the mirror was rotated θ degrees to the vertical, that β = θ. However,I haven't been able to show that γ = 2θ from the below diagram, as I am unable to determine the top right/bottom left angles of the parallelogram, or any additional angles of the triangles that exist in side of it.

Any assistance or guidance would be very much appreciated.

1. Homework Statement


If a laser beam is incident on a mirror, and the mirror is rotated θ degrees, prove that the angle by which the beam is deflected is equal to 2θ.

Homework Equations


[/B]
- Elementary geometric equations (e.g. sum of the interior angles of a triangle, properties of a parallelogram)

The Attempt at a Solution



YWlu2hN.jpg
 
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Kumo said:
Hi Everyone,

A question in my latest Physics I lab wishes for a proof showing that if a laser beam is incident on a mirror that is then rotated an angle θ the beam is deflected an angle 2θ.

I attempted to prove this geometrically below. In the diagram the angle γ, is the angle of deflection, and β is the incident angle. I was able to show that if the mirror was rotated θ degrees to the vertical, that β = θ. However,I haven't been able to show that γ = 2θ from the below diagram, as I am unable to determine the top right/bottom left angles of the parallelogram, or any additional angles of the triangles that exist in side of it.

Any assistance or guidance would be very much appreciated.

1. Homework Statement


If a laser beam is incident on a mirror, and the mirror is rotated θ degrees, prove that the angle by which the beam is deflected is equal to 2θ.

Homework Equations


[/B]
- Elementary geometric equations (e.g. sum of the interior angles of a triangle, properties of a parallelogram)

The Attempt at a Solution



YWlu2hN.jpg
How are the angles of incidence and reflection related?
 
SammyS said:
How are the angles of incidence and reflection related?

The angle of incidence is equal to the angle of reflection.
 
Kumo said:
The angle of incidence is equal to the angle of reflection.
Did you use that anywhere?

β = γ
 
SammyS said:
Did you use that anywhere?

β = γ


I see. That is what I initially thought, but then why is β ≠ θ when geometrically they seem to be equal when accounting for the isosceles triangle at the bottom of the diagram?
 
Kumo said:
I see. That is what I initially thought, but then why is β ≠ θ when geometrically they seem to be equal when accounting for the isosceles triangle at the bottom of the diagram?
β does equal θ
 
SammyS said:
β does equal θ

I see. Then given that β = γ = θ ≠ 2θ would that then mean that the angle of deflection is in fact not equal to twice the angle through which the mirror is rotated? Could it be a typo perhaps?
 
Kumo said:
I see. Then given that β = γ = θ ≠ 2θ would that then mean that the angle of deflection is in fact not equal to twice the angle through which the mirror is rotated? Could it be a typo perhaps?
What are you calling the angle of deflection?
 
SammyS said:
What are you calling the angle of deflection?
γ
 
  • #10
Kumo said:
γ
No.

What is the angle between the incident ray and the reflected ray ?
 
  • #11
SammyS said:
No.

What is the angle between the incident ray and the reflected ray ?

Uh, I see. This makes sense -- the reflected angle = β + γ. Given that θ = β = γ, the deflected angle = 2β = 2θ. Thank you very much for your help!
 
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