Draw a ray diagram to show the formation of real image of the same size as that of the object placed in front of a converging lens. Using this ray diagram establish the relation between u, v and f for this lens.
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Thin Lens Formula: Suppose an object AB of finite size is placed normally on the principal axis of a thin convex lens (fig.). A ray AP starting from A parallel to the principal axis, after refraction through the lens, passes through the second focus F. Another ray AC directed towards the optical centre C of the lens, goes straight undeviated. Both the rays meet at A′ Thus A′ is the real image of A. The perpendicular A′ B′ dropped from A′ on the principal axis is the whole image of AB.
Let distance of object AB from lens = u Distance of image A′B′ from lens = v Focal length of lens = f. We can see that Triangles ABC and A′B′C′ are similar $\frac{\text{AB}}{\text{A}'\text{B}'}=\frac{\text{CB}}{\text{C}'\text{B}'}\dots(\text{i})$ Similarly triangles PCF and A'B'F are similar $\frac{\text{P}\text{C}}{\text{A}'\text{B}'}=\frac{\text{CF}}{\text{F}'\text{B}}$ But PC = AB $\frac{\text{AB}}{\text{A}'\text{B}'}=\frac{\text{CF}}{\text{FB}'}\dots(\text{ii})$ From (i) and (ii), we get $\frac{\text{CB}}{\text{CB}'}=\frac{\text{CF}}{\text{FB}'}\dots(\text{iii})$ From sign convention, CB = -u, CB' = v, CF = f and FB' = CB' - CF = v - f Substituting this value in (iii), we get, $-\frac{\text{u}}{\text{v}}=\frac{\text{f}}{\text{v}-\text{f}}$ or -u (v - f) = vf or -uv + uf = uf Dividing throughout by uvf, we get $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\dots(\text{iv})$
Let distance of object AB from lens = u Distance of image A′B′ from lens = v Focal length of lens = f. We can see that Triangles ABC and A′B′C′ are similar $\frac{\text{AB}}{\text{A}'\text{B}'}=\frac{\text{CB}}{\text{C}'\text{B}'}\dots(\text{i})$ Similarly triangles PCF and A'B'F are similar $\frac{\text{P}\text{C}}{\text{A}'\text{B}'}=\frac{\text{CF}}{\text{F}'\text{B}}$ But PC = AB $\frac{\text{AB}}{\text{A}'\text{B}'}=\frac{\text{CF}}{\text{FB}'}\dots(\text{ii})$ From (i) and (ii), we get $\frac{\text{CB}}{\text{CB}'}=\frac{\text{CF}}{\text{FB}'}\dots(\text{iii})$ From sign convention, CB = -u, CB' = v, CF = f and FB' = CB' - CF = v - f Substituting this value in (iii), we get, $-\frac{\text{u}}{\text{v}}=\frac{\text{f}}{\text{v}-\text{f}}$ or -u (v - f) = vf or -uv + uf = uf Dividing throughout by uvf, we get $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\dots(\text{iv})$
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