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Question
Derivation of mirror formula.
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Solution
Mirror Formula:
The mirror formula is used to give the relation between the focal length of the mirror, the object O and the image V. The mirror formula for the concave mirror is given by;
1f=1V+1U;
Assumptions used in the derivations:
1. Distances are being measured from the pole of the mirror;
2. As per the convention positive sign indicates distance measured in the direction of incident rays while a negative sign indicates distance measured in the direction opposite to that of incident rays.
3. Also, the distance above the axis is positive while below is negative.
Consider the diagram as shown below:
From the figure, it is clear that the object AB has been placed at a distance of U from P (pole of the mirror). We can also deduce from the figure that the image A1B1 is formed at V from the mirror.
Now from the diagram it is clear that the opposite angle ACB and A1CB1 are equal ( law of vertically opposite angles).
∠ACB=∠A1CB1;
Similarly;
∠ABC=∠A1B1C; (right angles)
Now since two angles of triangle ACB and A1CB1 are equal and hence the third angle is also equal and is given by;
∠BAC=∠B1A1C; and
ABA1B1=BCB1C;...1
Similarly the triangle of FED and FA1B1 are also equal and similar, so;
EDA1B1=EFFB1;
Also since ED is equal to AB so we have;
ABA1B1=EFFB1;...2
Combining 1 and 2 we have;
BCB1C=EFFB1
Consider that the point D is very close to P and hence EF = PF, so;
BCB1C=PFFB1;
From the above diagram BC=PC−PB and B1C=PB1−PC and FB1=PB1−PF;
(PC−PB)PB1−PC=PF(PB1−PF);
Now substituting the values of above segments along with the sign, we have;
PC = -R;
PB = u;
PB1=−v;
PF = -f;
So the above equation becomes;
{−R−(−u)}{−v−(−R)}=−f{−v−(−f)}
(u−R)R−v=−f(f−v);
(u−R)(R−v)=f(v−f);
Solving it we have;
uv - uf - Rv + Rf = Rf - vf;
uv - uf - Rv + vf = 0;
since R = 2f (radius of curvature is twice that of focal length), hence;
uv - uf - 2fv + vf = 0;
uv - uf - vf = 0;
Solving it further and dividing with "uv" we have;
1f=1V+1U.
The mirror formula is used to give the relation between the focal length of the mirror, the object O and the image V. The mirror formula for the concave mirror is given by;
1f=1V+1U;
Assumptions used in the derivations:
1. Distances are being measured from the pole of the mirror;
2. As per the convention positive sign indicates distance measured in the direction of incident rays while a negative sign indicates distance measured in the direction opposite to that of incident rays.
3. Also, the distance above the axis is positive while below is negative.
Consider the diagram as shown below:
From the figure, it is clear that the object AB has been placed at a distance of U from P (pole of the mirror). We can also deduce from the figure that the image A1B1 is formed at V from the mirror.
Now from the diagram it is clear that the opposite angle ACB and A1CB1 are equal ( law of vertically opposite angles).
∠ACB=∠A1CB1;
Similarly;
∠ABC=∠A1B1C; (right angles)
Now since two angles of triangle ACB and A1CB1 are equal and hence the third angle is also equal and is given by;
∠BAC=∠B1A1C; and
ABA1B1=BCB1C;...1
Similarly the triangle of FED and FA1B1 are also equal and similar, so;
EDA1B1=EFFB1;
Also since ED is equal to AB so we have;
ABA1B1=EFFB1;...2
Combining 1 and 2 we have;
BCB1C=EFFB1
Consider that the point D is very close to P and hence EF = PF, so;
BCB1C=PFFB1;
From the above diagram BC=PC−PB and B1C=PB1−PC and FB1=PB1−PF;
(PC−PB)PB1−PC=PF(PB1−PF);
Now substituting the values of above segments along with the sign, we have;
PC = -R;
PB = u;
PB1=−v;
PF = -f;
So the above equation becomes;
{−R−(−u)}{−v−(−R)}=−f{−v−(−f)}
(u−R)R−v=−f(f−v);
(u−R)(R−v)=f(v−f);
Solving it we have;
uv - uf - Rv + Rf = Rf - vf;
uv - uf - Rv + vf = 0;
since R = 2f (radius of curvature is twice that of focal length), hence;
uv - uf - 2fv + vf = 0;
uv - uf - vf = 0;
Solving it further and dividing with "uv" we have;
1f=1V+1U.
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