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Draw a ray diagram to show the formation of the image of an object placed between the optical centre and principal focus of a convex lens. Deduce the relationship between the object distance, image distance and focal length under the conditions stated. |
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Answer» Solution :The image formation of a linear object AB placed between the OPTICAL centre C and principal focus F. of a convex lens has been shown in Fig. 9.63. The image A.B. is VIRTUAL, erect and magnified. As `DeltaA.B.C` and `DeltaABC` are similar. Hence, `(A.B.)/(AB) = (CB.)/(CB)` Again `triangleA.B.F.` and `triangleLCF` are similar, hence, `(A.B.)/(LC) = (B.F)/(CF)` or `(A.B.)/(AB) = (B.F)/(CF)`......(II) `[therefore LC = AB]`  comparing (i) and (ii) we get `(CB.)/(CB) = (B.F)/(CF) = (CB. + CF)/(CF)` As per sing convention `CB = -u, CB. =-v` and `CF = +f` `therefore (-v)/(-u) = (-v+f)/f` or `-VF = uv - uf` Dividing both sides by UVF, and on rearranging, we get `1/v -1/u =1/f`, which is the requisite lens formula. |
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