Proving Optical Law in plane mirror

[rktpro]

I tried to prove that distance of image is same as the distance of object in a plane mirror.
The image attached is the ray diagram.

First of all, I have assumed object and image to be parallel to mirror.

First I proved angle TSA = angle TSa

Then I have made similar triangle TSA and triangle TSa with AA similarity. ( where angle ATS = angle aTS ----each 90)

Therefore, TS/TS = TA/Ta
=> TA=Ta ( TS/TS = 1 )
=> AS=aS
( corresponding parts of similar triangle are equal in ratio)
Using this, I have made triangle ABS and abs congruent.
where,
As=aS
angle ASB = angle aSB
angle ABS = angle aBS

=> AB=aB ( corresponding parts of congruent triangles are equal)

Therefor, size of image is same as that of object.

Also, Bs=bS
(corresponding parts of congruent triangles)

That is, distance of image from mirror is same as that of object from mirror.

Now, I want to know how to prove that angle of image is same as that of object with principal axis? Because in my above method I already assumed both to be 90 degree. In other words, how to prove that image is erect too?

 

[chrisbaird]

If you want to know the right way, you start with the general electromagnetic wave expression. Write it down for the incident, reflected, and transmitted waves, then apply boundary conditions at the surface separating free space from the material. Because the incident and reflected waves exist in the same medium and must obey appropriate boundary conditions at all points on the planar interface, you find that they have the same frequency, same wave number, and same angle to the normal (on opposite sides).

 

[Stonebridge]

I tried to prove that distance of image is same as the distance of object in a plane mirror.
The image attached is the ray diagram.

First of all, I have assumed object and image to be parallel to mirror.

First I proved angle TSA = angle TSa

Then I have made similar triangle TSA and triangle TSa with AA similarity. ( where angle ATS = angle aTS ----each 90)

Therefore, TS/TS = TA/Ta
=> TA=Ta ( TS/TS = 1 )
=> AS=aS
( corresponding parts of similar triangle are equal in ratio)
Using this, I have made triangle ABS and abs congruent.
where,
As=aS
angle ASB = angle aSB
angle ABS = angle aBS

=> AB=aB ( corresponding parts of congruent triangles are equal)

Therefor, size of image is same as that of object.

Also, Bs=bS
(corresponding parts of congruent triangles)

That is, distance of image from mirror is same as that of object from mirror.

Now, I want to know how to prove that angle of image is same as that of object with principal axis? Because in my above method I already assumed both to be 90 degree. In other words, how to prove that image is erect too?

You have actually already proved it.
You have proved that the image of point A is at the point marked a and that this point is
-behind the mirror
-equidistant
-in the position shown

Now do the same for any other point on the object and you will find the same result.
All points on the object have their image equidistant behind the mirror.
This means point B has an image at b.
Therefore the object as a whole is exactly where you placed it.

 

[rktpro]

chrisbaird, I am unaware of this physics of waves. I am going to get a read at it. But, I am very thankful to you that this can be verified.

Stonebridge, The proof by assumption may be wrong. I wanted to know why the angles are same. I have proved it all using this assumption.