Ray reflecting off intersecting mirrors

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The solution is,

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]

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.

 

[ChiralSuperfields]

Try drawing it with θ very small.

Thank you for your reply @haruspex!

Like this?

Many thanks!

 

[haruspex]

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?

 

[ChiralSuperfields]

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]

Using the law of reflection

Which states …?

 

[ChiralSuperfields]

Thank you for your reply @haruspex!

Angle of incidence is equal to angle of reflection

Many thanks!

 

[haruspex]

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.

 

[ChiralSuperfields]

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!