Plane Mirror Image distance


Take a look at the animation that is
being played. In this animation a chess board has been taken and a king piece.
Now you will notice that as the king piece is being moved, the image of the King
piece is moving along with it. You will see that if the King piece is kept
two squares from the mirror, the image is also kept two squares from the mirror. If
the King piece is moved absolutely to the back of the chessboard the image of
the King piece moves the back of the chessboard. So how do you think the distance of the
king piece from the mirror is related to the distance of the image from the mirror?
Let us find out. In a plane mirror, is the distance of the
image from the mirror always equal to the distance of the object from the
mirror? Here we are going to prove that these two distances are equal with the
help of ray diagram. Now, consider the diagram that being shown on the board. Here
O is the object, I is the image being formed by the object
and this denoted by P is the plane mirror Now here we are going to consider two rays,
one incident ray is going to be incident Normally, since this wave is incident normally
the angle of incidence is 0, so according to the laws of reflection angle of
reflection will be zero. So it will be reflected back along its path and we are
going to consider another ray that is incident at an angle I which after reflection moves in this direction. So the two rays
that we are considering are OD and OB. As you will notice that OD is the incident
which after reflection is reaching its path and OB is an incident which after
reflection is travelling as BA. The observer is placed at A. Here in order to obtain
the image we are extrapolating these two rays and getting the image I. Now we consider that the angle of
Incidence is theta. So according to the laws of reflection the angle of reflection will also be
theta. Notice that BN is the normal at the point of the
incidence which means BN is perpendicular to the plane mirror. So angle
DBN is ninety degrees. This angle is ninety degrees. So
we can say that angle DBO is ninety minus theta. Similarly, angle DBM is also ninety degrees and now notice, angle NBA that is the angle of
reflection, it is theta Now, angle NBA is vertically opposite to angle MBI
and vertically opposite angles are always equal, so angle MBI is also theta. So that makes angle DBI 90-theta. So, here we have the angles as I just
Mentioned. Now we have to prove that the object
Distance that is OD is equal to the image distance that is ID. Now consider the diagram. Here in order to obtain our proof we have to
consider two triangles, triangle BOD and triangle BID that is triangle BOD an triangle BID. In these two triangles as I mentioned, angle BDO that is this angle and anger BDI, this
angle, are equal as both are ninety degrees. Angle OBN that is, this angle, is equal to angle
ABN, which is this angle, as angle of incidence is always equal to the angle
of reflection And thirdly angle ABN, that is this angle ABN, is equal to angle IBM, that is this angle is equal to this angle as
they are vertically opposite angles Now we already know that MN is
perpendicular to the mirror. Why? Because it is the normal at the point of incidence. So what can we conclude now? We can say that angle OBD and angle IBD are
Equal. Why are they equal? Recall as I mentioned that angle DBN is equal to angle DBM. Both of these are ninety degrees. Now we
know that angle OBN is theta, the angle of incidence, so if OBN is
theta and angle DBN is ninety, so angle DBO or angle OBD will be 90-theta. Similarly, we
can say that angle DBI will also be 90-theta because this angle that is angle IBM is theta and angle DBM is ninety degree, thus angle DBI will be
ninety minus theta. So these two angles are equal We can also say that BD is Equals to BD
Why? because line BD is the line that is common to both the triangles. And lastly, we can say that angle DOB is equal to
angle DIB, that is, this angle is equal to this angle. Why? Because as you know the sum of
three angles in a triangle is always equal to a hundred and eighty degrees, so
if you need to find the value of this Angle, we need to subtract the sum of
these two from 180 so that will be 180 minus, the value of
this angle is 90 plus the value of this angle, that is
90-theta. This gives us theta because 180 and 180 cancel out. So angle BOD is theta. Similarly,
in triangle DBI, if we consider the sum of angle DBI and angle BDI, we come up
with a similar calculation and thus we can say that angle DIB is equal to theta. So, angle DOB is equal to angle DIB.
Now as we can see, we have a pair of angles that are equal, a side that is common and another pair of
angles that are equal. So what can we say? We can say the triangle BOD, that is triangle BOD, is congruent
to triangle BID through angle side angle congruency or ASA congruency. Now since these
two triangles are congruent, we can say that side OD is equal to side ID
because corresponding parts of congruent triangles
are equal So, we can see that OD is equal to ID,
and what was OD? OD was the object distance and what was ID? ID with the
image distance. So what can we conclude? We can thus conclude that for a plane mirror, the distance of the object from the mirror
is always equal to the distance of the image from the mirror.

29 Replies to “Plane Mirror Image distance

  1. i think u should have used BDO=BDI=90 and then asa would hold true cus the way u hv done the two angles dont have a common side

  2. Actually I didn't understand anything about this in school ,but when I saw this I understood perfectly 100%

  3. You done a little mistake in writing congruence creteria,you write ASA but it is wrong triangle is congruence by AAS creteria.
    Other part is good.

  4. the congruency criteria u have used is wrong. Other part is well explained. U all have to work on mathematics, IIM graduates.

  5. Video was nice but there is a mistake , u mentioned that triangles are congrequent by asa but actually a a s

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