Two transparent media having refractive indic | vedclass

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(A) Given: Refractive index of rarer medium $\mu_1 = 1.0$,refractive index of denser medium $\mu_2 = 1.5$. Radius of curvature $R = +30\,cm$ (as the c

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$30$

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(A) Given: Refractive index of rarer medium $\mu_1 = 1.0$,refractive index of denser medium $\mu_2 = 1.5$. Radius of curvature $R = +30\,cm$ (as the centre of curvature is in the direction of light propagation). Object distance $u = -15\,cm$ (measured against the direction of light).Using the formula for refraction at a spherical surface:$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$Substituting the values:$\frac{1.5}{v} - \frac{1.0}{-15} = \frac{1.5 - 1.0}{30}$$\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$$\frac{1.5}{v} + \frac{1}{15} = \frac{1}{60}$$\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15}$$\frac{1.5}{v} = \frac{1 - 4}{60} = \frac{-3}{60} = -\frac{1}{20}$$v = 1.5 	imes (-20) = -30\,cm$.The negative sign indicates that the image is formed on the same side as the object at a distance of $30\,cm$ from the pole.

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