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(C) For refraction at a spherical surface,the formula is $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Here,the object is at the center,so $u

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$6$

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(C) For refraction at a spherical surface,the formula is $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Here,the object is at the center,so $u = -R = -6 \, cm$.The light travels from glass $(n_1 = 1.5)$ to air $(n_2 = 1.0)$.The radius of curvature $R$ for the surface is $-6 \, cm$ (as the center of curvature is to the left of the surface).Substituting the values: $\frac{1.0}{v} - \frac{1.5}{-6} = \frac{1.0 - 1.5}{-6}$.$\frac{1}{v} + 0.25 = \frac{-0.5}{-6} = \frac{0.5}{6} = \frac{1}{12}$.$\frac{1}{v} = \frac{1}{12} - \frac{1}{4} = \frac{1 - 3}{12} = \frac{-2}{12} = -\frac{1}{6}$.Thus,$v = -6 \, cm$.The negative sign indicates that the image is virtual and formed on the same side as the object,at a distance of $6 \, cm$ from the surface.

$A$ point object is placed at the centre of a glass sphere of radius $6 \, cm$ and refractive index $1.5$. The distance of the virtual image from the surface of the sphere is.....$cm$.

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