A point object O is placed in front of a glas | vedclass

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(A) Given: Object distance $u = -15 \,cm$ (as it is placed to the left of the pole).Radius of curvature $R = +30 \,cm$ (as the center of curvature is

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$30 \,cm$ to the left

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(A) Given: Object distance $u = -15 \,cm$ (as it is placed to the left of the pole).Radius of curvature $R = +30 \,cm$ (as the center of curvature is to the right of the pole).Refractive index of air $\mu_1 = 1$.Refractive index of glass $\mu_2 = 1.5$.Using the refraction formula 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}{-15} = \frac{1.5 - 1}{30}$$\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$$\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 $30 \,cm$ to the left of the pole,on the same side as the object.

$A$ point object $O$ is placed in front of a glass rod having a spherical end of radius of curvature $30 \,cm$. The image would be formed at

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