$A$ point object $O$ is placed on the axis of a cylindrical piece of glass of refractive index $1.6$ as shown in the figure. One surface of the glass piece is convex with a radius of curvature $3 \,mm$. The point appears to be at $5 \,mm$ on the axis when viewed along the axis from the right side of the convex surface. The distance of the point object from the convex surface is: (in $\,mm$)

$A$ transparent thin film of uniform thickness and refractive index $n_1=1.4$ is coated on the convex spherical surface of radius $R$ at one end of a long solid glass cylinder of refractive index $n_2=1.5$,as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance $f_1$ from the film,while rays of light traversing from glass to air get focused at distance $f_2$ from the film. Then:
$(A)$ $|f_1|=3R$
$(B)$ $|f_1|=2.8R$
$(C)$ $|f_2|=2R$
$(D)$ $|f_2|=1.4R$

$A$ 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 $R$)

$A$ spherical glass is attached to a rigid wall as shown in the figure. An observer located at point $O$ is looking at a point $A$ on the wall. The refractive index of the glass is $1.5$ and that of air is $1.0$. The distances are $OA = 8 \text{ cm}$, $XA = 3 \text{ cm}$. If the radius of curvature of the spherical glass surface is $R = 5 \text{ cm}$, then the apparent distance of $A$ from the observer $O$ is (in $\text{ cm}$)

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 $; cm$)?

$A$ spherical surface of radius of curvature $R$ separates air (refractive index $1.0$) from glass (refractive index $1.5$). The centre of curvature is in the glass. $A$ point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $O$,and $PO = OQ$. The distance $PO$ is equal to (in $R$)