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Derive the lens formula 1/f = 1/v -1/ufor a concave lens, using the necessary ray diagram. |
Answer» Solution :The image formation of an object AB PLACED in front of a thin CONCAVE lens has been shown in Fig. 9.64. The image A.B. is a virtual, erect and diminished image formed between F and C. As AA.B.C and AABC are similar, hence  `(A.B.)/(AB) = (CB.)/(CB)`.......(i) Again as `triangleA.BF` and `triangleLCF` are similar, hence `(A.B.)/(LC) = (B.F)/(CF)` or `(A.B.)/(AB) = (B.F)/(CF)`.......(ii) `[THEREFORE LC = AB]` Comparing (ii) and (i), we get `(CB.)/(CB) = (B.F)/(CF) = (CF - CB.)/(CF)` As per sign convention being followed, `CB = -u, CB. = -V` and `CF = -f` `therefore (-v)/(-u) = ((-f)-(-v))/(-f)` or `vf = uf - uv` Dividing both sides by uvf, and rearranging the terms, we get `1/v - 1/u = 1/f`, which is the requisite lens formula. |
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