Ray reflecting off intersecting mirrors

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Homework Help Overview

The discussion revolves around the behavior of light rays reflecting off intersecting mirrors, particularly focusing on the conditions under which multiple reflections occur and the implications of the angle of incidence.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants express confusion regarding the conditions for multiple reflections, specifically questioning the meaning of a negative angle, denoted as ## \gamma < 0##. There are attempts to visualize the problem by suggesting drawing diagrams with small angles. Discussions also include the application of the law of reflection and the correct interpretation of angles in relation to the normal of the mirror.

Discussion Status

Participants are actively engaging with each other's responses, seeking clarification on the reflections and the angles involved. Some have provided guidance on how to properly draw the reflections and the normals, while others are in the process of revising their diagrams based on feedback.

Contextual Notes

There is mention of the diagrams not being to scale, which may affect the understanding of the reflections. Participants are also navigating the complexities of visualizing the angles and reflections accurately.

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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!
 
Last edited by a moderator:
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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