Answer:
Assumptions used in the derivation of lens formula: (i) The lens used in thin. (ii) The aperture of the lens is small. (iii) The incident and refracted rays make small angles with the principal axis. (iv) The object is a small object placed on the principal axis. Derivation of thin lens formula for a convex lens when it forms a real image. As shown in Fig. consider an object AB placed perpendicular to the principal axis of thin convex lens between its F? and C?. A real, inverted and magnified image A?B? is formed beyond C on the other side of the lens. \[\Delta A'B'O\,and\,\Delta ABO\,are\,similar,\] \[\therefore \] \[\frac{A'B'}{AB}=\frac{OB'}{BO}\] ?(1) \[Also\,\Delta A'B'F\,and\,\Delta MOF\,are\,similar,\] \[\therefore \] \[\frac{A'B'}{MO}=\frac{FB'}{OF}\] \[But\,MO=AB,\] \[\therefore \] \[\frac{A'B'}{AB}=\frac{FB'}{OF}\] ?(2) From (1) and (2), we get \[\frac{OB'}{BO}=\frac{FB'}{OF}=\frac{OB'-OF}{OF}\] Using new Cartesian sign convention, we get Object distance, \[BO=-u\] Image distance, \[OB'=+\upsilon \] Focal length, \[OF=+f\] \[\therefore \] \[\frac{\upsilon }{-u}=\frac{\upsilon -f}{f}\] or \[\upsilon f=-u\upsilon +uf\,\,or\,\,u\upsilon =uf-\upsilon f\] Dividing both sides by \[u\upsilon f,\] we get \[\frac{1}{f}=\frac{1}{\upsilon }-\frac{1}{u}\] This proves the lens formula for a convex lens when it forms a real image. The linear magnification produced by a lens is define as the ratio of the size of the image (h\[({{h}_{2}})\] formed by the lens to the size of the object \[({{h}_{1}})\]. \[m=\frac{{{h}_{2}}}{{{h}_{1}}}=\frac{\upsilon }{u}\] The graph drawn between linear magnification \[m\] an the image distance \[\upsilon \] is a straight line as shown in Fig. By reading the values of \[\text{m}\] and \[\upsilon \] from this graph, we can find the focal length \[f\] of the lens from the following relation: \[m=\frac{f-\upsilon }{f}\]
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