The graph between frac{1}{u} and frac{1}{v} f | vedclass

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(B) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.From the graph,at point $A$,$\frac{1}{u} = 0$ and $\frac{1}{v} = 0.10 \, cm

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$10$

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(B) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.From the graph,at point $A$,$\frac{1}{u} = 0$ and $\frac{1}{v} = 0.10 \, cm^{-1}$. Substituting these into the lens formula: $0.10 - 0 = \frac{1}{f} \Rightarrow f = 10 \, cm$.At point $B$,$\frac{1}{u} = -0.10 \, cm^{-1}$ and $\frac{1}{v} = 0$. Substituting these into the lens formula: $0 - (-0.10) = \frac{1}{f} \Rightarrow \frac{1}{f} = 0.10 \Rightarrow f = 10 \, cm$.Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a convex lens with equal radii of curvature,$R_1 = R$ and $R_2 = -R$.So,$\frac{1}{10} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R}$.Therefore,$R = 10 \, cm$.

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