Iinterference in thin films due to reflected light

Click For Summary

Homework Help Overview

The discussion revolves around the topic of interference in thin films, specifically focusing on the angles associated with incident, reflected, and refracted rays. The original poster seeks clarification on the definition and geometry of the angle of incidence in relation to a diagram provided.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions of angles in the context of reflection and question the geometry involved in identifying the angle of incidence. There is a focus on the relationships between various angles and the application of geometric theorems to support their reasoning.

Discussion Status

The discussion is active, with participants providing insights into geometric principles and engaging in clarifying the relationships between angles. Some participants express understanding while others seek further explanation on the basis for angle equality.

Contextual Notes

There is an emphasis on the need for clarity regarding the definitions of angles in the context of the provided diagram, and participants are navigating through potential misunderstandings related to angle terminology and geometry.

logearav
Messages
329
Reaction score
0

Homework Statement




dear revered members,
please find the attached image for the topic interference in thin films due to relection.

Homework Equations


my questions are
1) angle between the incident ray or reflected ray or refracted ray is called angle of incidence, angle of reflection and angle of refraction, as the case needs. so <AON is angle of incidence. how <ACN is angle of incidence? because CN is normal line and AC IS POINT OF CONTACT, then how it is angle of incidence?
is there any geometry involcd in it?
ANY HELP IN THIS REGARD WILL BE GREATLY APPRECIATED, REVERED MEMBERS


The Attempt at a Solution


 

Attachments

  • IMG_0001.jpg
    IMG_0001.jpg
    49.9 KB · Views: 476
Physics news on Phys.org
Yes, there is a theorem in Euclidean geometry that says that if two angles are less than 90o and their sides are mutually perpendicular, then the angles are equal. Label the top of the dotted line that is perpendicular to the interface as Q. Then angle SAQ is equal to QAT by the law of reflection. Note that line QA is perpendicular to AC and line AT is perpendicular to line NC.
 
sir, i understood the concept. but, in my attachment SA is incident ray and NA is the normal, so <SAN is angle of incidence. but, NC is normal to AT and CQ. <i is between AC and CN. how can it be termed as angle of incidence. because both AT and CQ, according to my attachment are reflected rays or transmitted rays. so they should be <r. Why <i there?
 
I am not disputing that SA is the incident ray. The angle of incidence is SAI. This angle in degrees is equal to the reflection angle INT in degrees and also equal to angle ACN in degrees. I never said that ACN is the angle of incidence; I said it is equal to the angle of incidence in degrees. Do you see the difference?
 
thanks you sir. but on what basis they are equal? adjacent angles? alternate angles? or something else.
 
kuruman said:
Yes, there is a theorem in Euclidean geometry that says that if two angles are less than 90o and their sides are mutually perpendicular, then the angles are equal. Label the top of the dotted line that is perpendicular to the interface as Q. Then angle SAQ is equal to QAT by the law of reflection. Note that line QA is perpendicular to AC and line AT is perpendicular to line NC.

They are equal on the basis shown above.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
994
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 35 ·
2
Replies
35
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
2K
Replies
2
Views
4K