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rktpro
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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?

post-33891-0-34055900-1317868782_thumb.png

Edited by rktpro
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  • 2 weeks later...

 

In other words, how to prove that image is erect too?

 

Sorry if no help. Not my field, but since nobody else was contributing I thought this video would perhaps put you on the right track. The video explains lateral inversion, but perhaps doesn't explain lack of vertical inversion. My apologies.

 

Apparently what you have asked is a surprisingly difficult problem (of which you might be well aware). You might like to read this link:- http://www.answers.com/topic/mirror-reversal

Edited by TonyMcC
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  • 1 month later...

Yes the distances will be the same because there is no magnification. And your picture should have had at least two rays, for the top and bottom of the object.

 

Because the object is on principal axis, the ray from bottom would retrace its path.

How can you show that there is no magnification? Isn't my way correct because it involved no steps which first proves no magnification?

The real question still remains unanswered, Sir.

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