$A$ laser beam falls on a crystal ball of radius $R$ as shown in the figure. What is its refractive index?

$A$ point source of light at the surface of a sphere causes a parallel beam of light to emerge from the opposite surface of the sphere. The refractive index of the material of the sphere is

$A$ clear transparent glass sphere $(\mu=1.5)$ of radius $R$ is immersed in a liquid of refractive index $1.25$. $A$ parallel beam of light incident on it will converge to a point. The distance of this point from the center will be

An air bubble in a glass sphere having a $4 \,cm$ diameter appears $1 \,cm$ from the surface nearest to the eye when looked at along the diameter. If the refractive index of glass is $_a\mu_g = 1.5$,the actual distance of the bubble from the refracting surface is.....$cm$.

$A$ spherical surface of radius of curvature $R$ separates air (refractive index $1.0$) from glass (refractive index $1.5$). The centre of curvature is in the glass. $A$ point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $O$,and $PO = OQ$. The distance $PO$ is equal to (in $R$)

$A$ concave spherical refracting surface separates two media,glass and air $(\mu_1 = 1.5, \mu_2 = 1.0)$. If the image is to be real,at what minimum distance $u$ should the object be placed in the glass if $R$ is the radius of curvature?