A lens is made of glass having an index of re | vedclass

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(A) Given: Refractive index $\mu = 1.5$, object distance $u = -60 \ cm$, image distance $v = +120 \ cm$.For a plano-convex lens, let the radius of the

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$20$

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(A) Given: Refractive index $\mu = 1.5$, object distance $u = -60 \ cm$, image distance $v = +120 \ cm$.For a plano-convex lens, let the radius of the convex surface be $R_1 = -R$ (as it faces the object) and the flat surface be $R_2 = \infty$.Using the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = 0.5 \left( -\frac{1}{R} \right) = -\frac{1}{2R}$So, $f = -2R$.Using the lens formula:$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\frac{1}{120} - \frac{1}{-60} = \frac{1}{-2R}$$\frac{1}{120} + \frac{1}{60} = -\frac{1}{2R}$$\frac{1 + 2}{120} = -\frac{1}{2R}$$\frac{3}{120} = -\frac{1}{2R} \Rightarrow \frac{1}{40} = -\frac{1}{2R}$$2R = -40 \ cm$. Since $R$ represents the magnitude of the radius, we take $R = 20 \ cm$.

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