$A$ ray coming from an object which is situated at infinity in the air falls on a spherical glass surface $(n=1.5)$. The distance of the image will be . . . . . . . ($R$ is the radius of curvature of the spherical glass surface.)

Parallel rays are incident on a transparent sphere along its one diameter. After refraction,these rays converge at the other end of this diameter. The refractive index of the sphere is:

Consider a point object $O$ situated at a distance of $30 \ cm$ from the centre of a sphere of radius $2 \ cm$ and refractive index $\mu_2 = 1.5$ as shown in the figure. If the refractive index of the region surrounding this sphere is $\mu_1 = 1.4$, then the position of the image due to refraction by the sphere with respect to the centre is:

$A$ 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?

Region $I$ and $II$ are separated by a spherical surface of radius $25\, \text{cm}$. An object is kept in region $I$ at a distance of $40\, \text{cm}$ from the surface. The distance of the image from the surface is $.....\, \text{cm}$.

$A$ spherical convex surface of power $5 \text{ dioptre}$ separates object and image space of refractive indices $1.0$ and $\frac{4}{3}$ respectively. The radius of curvature of the surface is (in $\text{ cm}$)