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.....$cm$.

$A$ 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:

$A$ parallel beam of light travelling in water (refractive index $\mu_1 = 4/3$) is refracted by a spherical air bubble of radius $R = 2 \, cm$ situated in water. Assuming the light rays to be paraxial,the position of the image due to refraction at the first surface is:

$A$ hemispherical glass lens with a refractive index of $1.5$ is placed in a liquid with a refractive index of $1.3$ (see the figure). The radius of the hemispherical lens is $10 \,cm$. $A$ parallel beam of light traveling in the liquid is refracted by the glass lens. The absolute value of the position of the image from the center of the glass lens will be (in $\,cm$)

In a long glass tube,a mixture of two liquids $A$ and $B$ with refractive indices $1.3$ and $1.4$ respectively,forms a convex refractive meniscus towards $A$. If an object placed at $13 \ \text{cm}$ from the vertex of the meniscus in $A$ forms an image with a magnification of $-2$,then the radius of curvature of the meniscus is:

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