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(C) Given: Radius of curvature $R = -10 \ cm$ (as the surface is concave towards the object in medium $X$),refractive index of medium $X$ (incident si

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image is always virtual

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(C) Given: Radius of curvature $R = -10 \ cm$ (as the surface is concave towards the object in medium $X$),refractive index of medium $X$ (incident side) $n_1 = 4/3$,and refractive index of medium $Y$ (refracted side) $n_2 = 3/2$.Let the object be placed at a distance $u = -x$ from the pole.Using the formula for refraction at a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Substituting the values: $\frac{3/2}{v} - \frac{4/3}{-x} = \frac{3/2 - 4/3}{-10}$.$\frac{1.5}{v} + \frac{1.333}{x} = \frac{0.1667}{-10} = -0.01667$.$\frac{1.5}{v} = -0.01667 - \frac{1.333}{x}$.Since $x$ is a positive distance,both terms on the right side are negative. Therefore,$v$ must be negative.$A$ negative value of $v$ indicates that the image is formed on the same side as the object,which means the image is always virtual.

$A$ concave spherical surface of radius of curvature $10 \ cm$ separates two media $X$ and $Y$ of refractive index $4/3$ and $3/2$ respectively. If the object is placed along the principal axis in medium $X$,then:

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