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.

$A$ 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:

$A$ small object is enclosed in a transparent solid sphere of radius $8 \ cm$. The object is situated at $2 \ cm$ from the centre of the sphere. If its image appears to be at $3.2 \ cm$ from the nearest side,then the refractive index of the material of the sphere is:

$A$ 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:
$(A)$ $|f_1|=3R$
$(B)$ $|f_1|=2.8R$
$(C)$ $|f_2|=2R$
$(D)$ $|f_2|=1.4R$

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:

$A$ spherical surface of radius of curvature $10\,cm$ separates two media $X$ and $Y$ of refractive indices $3/2$ and $4/3$ respectively. The centre of the spherical surface lies in the denser medium. An object is placed in medium $X$. For the image to be real,the object distance must be: