Why can γ be negative in the ray reflection formula?

member 731016 · Feb 25, 2023

For this problem,

Diagram of two flat mirrors meeting at angle θ with a light ray reflecting off the slanted mirror, then the horizontal mirror, and emerging at angle β between the emerging and incident rays.

The solution is,

Ray diagram of a light ray reflecting off Mirror 1 (horizontal) and Mirror 2 (tilted), forming triangle abc with angles α, γ, β and labels a, b, c, d, θ, 90°−α, 90°−γ; below it the equation 2α + 2γ + β = 180°.

Worked solution text with equations: β = 180° − 2(α + γ) [1]; (90.0° − α) + (90.0° − γ) + θ = 180°, giving θ = α + γ [2]; substituting gives β = 180° − 2θ; note that γ = θ − α and multiple reflections occur when α > θ.

I don't understand how if ## \gamma < 0## then there will be multiple reflections? I don't understand how ##\gamma## can be negative.

Many thanks!

haruspex · Feb 25, 2023

Callumnc1 said:

I don't understand how if ## \gamma < 0## then there will be multiple reflections? I don't understand how ##\gamma## can be negative.

Many thanks!

Try drawing it with θ very small.

member 731016 · Feb 25, 2023

haruspex said:

Try drawing it with θ very small.

Thank you for your reply @haruspex!

Like this?

Hand-drawn ray diagram in red: a light ray reflects off the lower surface of a wedge at angle θ, hits the upper surface, and exits to the right at angle β to the normal.

Many thanks!

haruspex · Feb 25, 2023

Callumnc1 said:

Thank you for your reply @haruspex!

Like this?
View attachment 322878
Many thanks!

First reflection looks fine, the second impossible. How do you figure out where a reflected ray goes?

member 731016 · Feb 26, 2023

haruspex said:

First reflection looks fine, the second impossible.

Thank you for your reply @haruspex!

How do you figure out where a reflected ray goes?

Using the law of reflection

EDIT: Apologies, the diagram is not to scale. Let me know if you want me to redo it and I will use a better program (Microsoft paint instead of snip tool)

Many thanks!

haruspex · Feb 26, 2023

Callumnc1 said:

Using the law of reflection

Which states …?

member 731016 · Feb 26, 2023

Thank you for your reply @haruspex!

Angle of incidence is equal to angle of reflection

Many thanks!

haruspex · Feb 26, 2023

Callumnc1 said:

Thank you for your reply @haruspex!

Angle of incidence is equal to angle of reflection

Many thanks!

But what you drawn for the second reflection is nothing like that.
Draw the normal to the mirror where the ray hits it. The incident ray makes some angle theta to that normal. The reflected ray should also make angle theta to the normal, but on the other side of the normal.

member 731016 · Feb 27, 2023

haruspex said:

But what you drawn for the second reflection is nothing like that.
Draw the normal to the mirror where the ray hits it. The incident ray makes some angle theta to that normal. The reflected ray should also make angle theta to the normal, but on the other side of the normal.

Thank you for your reply @haruspex!

I will draw another diagram. Sorry if my replies are a bit slow

Many thanks!

Why can γ be negative in the ray reflection formula?

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