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Answers
The formula is a relation between the object distance u , inmage distance v and the focal length from the pole of the concave mirror. The formula is valid for the images in convex mirror and even for the images in lens.
We consider the image formed by aconcave mirror whose focal length is f and whose radius of curvature is r = 2f.
Let P be the pole of the concave mirror. Let P, F , C be the pole, focucal point , and centre of curvature along principal axis . So, PC = 2PF , as r = 2f.
Let AB be a vertically standinding object beyond C on the principal axis.
Then the ray starting from B parallel to principal axis incident on the mirror at D reflects through the focus F. Let the reflected ray be CFB' .
The another ray starting from B through the centre C incident on the mirror at E retraces its path by reflection being normal to the mirror.
Now BE and DF produced meet at B'.
Now drop the perpendicular from B' to PC to meet at A'.
Drop the perpendicular from D to PC to meet at G.
Now PF = f , the focal length. PA = u object distance from the mirror. PA' = v the image distance.
Now consider the similar triangles ABC and A'BC.
AB/AB' = AC/ A'C =( PU-PC)(PC-PA') = (u-2f)/((2f-v).....(1)
Consider the similar triangles DFG and A'B'F.
DG/A'B' = PF/PA' PF/(PA'-PF)= f/(v-f)... (2)
DG = AB. So (2) could be rewritten as:
AB/A'B' + f/v ....................(3).
From (2) and (3), LHS being same , we can equate right sides.
(u-2f)/(2f-v) = f/(v-f).
(u-2f)(v-f) = (2f-v)f.
uv-2fv -fu +2f^2 = 2f^2 -fv
uv = fu +fv
Dvide by uvf;
1/f = 1/v+1/u.
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Answer:
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Explanation:
Derivation of formula for convex mirror: �Let AB be an object placed on the principal axis of a convex mirror of focal length f. u is the distance between the object and the mirror and v is the distance between the image and the mirror.� convex mirror formula In ABC and A1B1C <ABC = <A1B1C (right angles) <ACB = <A1CB1 <CAB = <CA1B (common angle) ABC is similar to A1B1C AB/A1B1 = BC/B1C........(1) similarly DEF issimilar to A1B1F DE/A1B1 = EF/B1F....(2) But DE = AB and when the aperture is very small EF = PF Equation (2) becomes� AB/A1B1 = PF/B1F....(3) Frm equations (1) and (3) get PF/B1F = BC/B1C PF/PF-PB1 = PB + PC/PC - PB1 f/f - v = -u + 2f/2f - v [PF = f, PB1 = v, PB = u, PC = 2f] 2(2f - v)= (f-v)(2f-u) i could not write the following 2 steps sorry -vf + uf + 2 fv -vu=0 fv+uf-vu=0....(4) Dividing both sides of equation(4) by uvf we get fv/uvf + uf/uvf - uv/uvf=0 1/u +1/v - 1/f=0 1/u + 1/v = 1/f