In a thin spherical fish bowl of radius $10 \, cm$ filled with water of refractive index $\mu = \frac{4}{3}$,there is a small fish at a distance of $4 \, cm$ from the center as shown in the figure. Where will the image of the fish appear if seen from point $F$ (in $, cm$)?

Refer to the figure given below. $\mu_1$ and $\mu_2$ are the refractive indices of air and the lens material,respectively. The height of the image will be . . . . . . cm.

$A$ quarter cylinder of radius $R$ and refractive index $1.5$ is placed on a table. $A$ point object $P$ is kept at a distance of $mR$ from it. Find the value of $m$ for which a ray from $P$ will emerge parallel to the table as shown in the figure.

$A$ point object is kept at $P$ in front of a glass sphere of radius $R$. Its image is formed at $Q$ such that $PO = QO$. The refractive index of the material of the glass sphere is $1.4$. The distance $PO$ is equal to:

$A$ parallel beam of light travelling in air (refractive index $1.0$) is incident on a convex spherical glass surface of radius of curvature $50 \ cm$. The refractive index of glass is $1.5$. The rays converge to a point at a distance $x \ cm$ from the centre of curvature of the spherical surface. The value of $x$ is . . . . . . $cm$.

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