Ray reflecting off intersecting mirrors

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The discussion revolves around understanding multiple reflections of light between intersecting mirrors, particularly addressing the confusion regarding negative angles (γ < 0) and their implications. Participants emphasize the importance of accurately applying the law of reflection, which states that the angle of incidence equals the angle of reflection. There is a request for clearer diagrams to illustrate the reflections, as initial attempts were deemed inaccurate. The conversation highlights the need for proper representation of angles and normals in the drawings to facilitate understanding. Overall, the focus is on clarifying the mechanics of light reflection in this geometric context.
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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1677384446103.png

The solution is,
1677384477316.png

1677384490654.png

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!
 
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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.
 
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haruspex said:
Try drawing it with θ very small.
Thank you for your reply @haruspex!

Like this?
1677388295033.png

Many thanks!
 
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?
 
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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!
 
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Callumnc1 said:
Using the law of reflection
Which states …?
 
Thank you for your reply @haruspex!

Angle of incidence is equal to angle of reflection

Many thanks!
 
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.
 
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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!
 
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