rktpro Posted October 6, 2011 Posted October 6, 2011 (edited) 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? Edited October 6, 2011 by rktpro 1
rktpro Posted October 7, 2011 Author Posted October 7, 2011 Anyone please reply. I am dying to learn it.
TonyMcC Posted October 18, 2011 Posted October 18, 2011 No one got an idea? I haven't done light ray diagrams since I was at school, but this link has a video that might provide the information you seek. 1
rktpro Posted October 19, 2011 Author Posted October 19, 2011 Have you seen the video Tony? It doesn't answer my question.
TonyMcC Posted October 19, 2011 Posted October 19, 2011 (edited) 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 October 19, 2011 by TonyMcC
Mr Skeptic Posted November 21, 2011 Posted November 21, 2011 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.
rktpro Posted November 21, 2011 Author Posted November 21, 2011 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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