The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea $(7.8 \, mm)$. This surface separates two media of refractive indices $1$ and $1.34$. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus in $cm$.

$A$ glass rod has ends as shown in the figure. The refractive index of glass is $\mu$. The object $O$ is at a distance $2R$ from the surface of larger radius of curvature. The distance between the apexes of the ends is $3R$. The range of $\mu$ for which the image is real is given by

$A$ spherical surface of radius of curvature $R$ separates air from glass (refractive index $= 1.5$). The centre of curvature is in the glass medium. $A$ point object $O$ is placed in air on the optic axis of the surface,so that its real image is formed at $I$ inside the glass. The line $OI$ intersects the spherical surface at $P$ and $PO = PI$. The distance $PO$ is equal to: (in $R$)

Two transparent media having refractive indices $1.0$ and $1.5$ are separated by a spherical refracting surface of radius of curvature $30\,cm$. The centre of curvature of the surface is towards the denser medium and a point object is placed on the principal axis in the rarer medium at a distance of $15\,cm$ from the pole of the surface. The distance of the image from the pole of the surface is .......$cm$.

$A$ curved surface of radius $R$ separates two media of refractive indices $\mu_1$ and $\mu_2$ as shown in figures $A$ and $B$. Identify the correct statement$(s)$ related to the formation of images of a real object $O$ placed at distance $x$ from the pole of the concave surface,as shown in figure $B$.

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: