An air bubble in a glass sphere $(\mu=1.5)$ is situated at a distance $3\; cm$ from a convex surface of diameter $10\; cm$. At what distance from the surface will the bubble appear (in $; cm$)?

An object is placed at a distance of $20 \, cm$ in a rarer medium from the pole of a convex spherical refracting surface of radius of curvature $10 \, cm$. If the refractive index of the rarer medium is $1$ and that of the denser medium is $2$,then the position of the image is at

$A$ 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 $R$)

$A$ parallel beam of light emerges from the opposite surface of the sphere when a point source of light lies at the surface of the sphere. The refractive index of the sphere is

Water (with refractive index $\mu = \frac{4}{3}$) in a tank is $18 \ cm$ deep. Oil of refractive index $\mu = \frac{7}{4}$ lies on the water,forming a convex surface of radius of curvature $R = 6 \ cm$ as shown. Consider the oil to act as a thin lens. An object $S$ is placed $24 \ cm$ above the water surface. The location of its image is at $x \ cm$ above the bottom of the tank. Then $x$ is

As shown in the figure,region $BCDEF$ and $ABFG$ have refractive indices $2.0$ and $1.5$ respectively. $A$ particle $O$ is kept at the midpoint of region $BCEF$. The final image of the object as seen by the eye is at a distance of: