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Derive mirror equation. |
Answer» Solution : MPN is a concave mirror of radius of curvature 2f and focal length f. AB represents an extended object and A.B. its image. For objects beyond C, the image is real and inverted and is formed in b/w F and C. The image will be diminished. Let u be the object distance from P and v be the image distance from P. Sign convention : (1) All distances aremeasured from the POLE of the mirror. (2) Distances measured from P and to the opposite direction of incident light are taken as negative. Let .M. be very close to P. Then the arc lengthMP = perpendicular length PM. Considerright angled triangles A.B.F and MPF `(A^(1)B^(1))/(MP)=(B^(1)F)/(FP)`, SINCE `MP=AB, (A^(1)B^(1))/(AB)=(B^(1)F)/(FP)""`...........(1) where,`B^(1)F=B^(1)P-FP=v-f "" `.........(2) Consider two other SIMILER rightangled triangles ABP and `A^(1)B^(1)P`. `(A^(1)B^(1))/(AB) = (B^(1)P)/(BP) = (-v)/(-u) "" ` .........(3) comparing (1) and (3) since`PF=-f, B^(1)P=-v`, be sign conventions we get, i.e., `(B^(1)F)/(FP)= (B^(1)P)/(BP)` i.e., `(v-f)/(f)= (v)/(u)` i.e., `(v)/(f)-1=(v)/(u) "" divide v` on both sides, we get`(1)/(f)-(1)/(v)=(1)/(u)` Hence, `(1)/(f)=(1)/(u)+(1)/(v)` This is known as the mirror formula. Note : Magnification,`m=(v)/(u). ""` By sign convention `h_(0)` is +ve, and for read image `h_(i)` is -ve. hence, `m=(-h_(i))/(h_(o))=(v)/(u)` or`m=(-v)/(u)` |
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