$A$ 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:

$A$ convex refracting surface of radius of curvature $20 \, cm$ separates two media of refractive indices $\frac{4}{3}$ and $1.6$. An object is placed in the first medium $(\mu = 4/3)$ at a distance of $200 \, cm$ from the refracting surface. The position of the image formed is.....$cm$

An air bubble in a glass sphere having a $4 \,cm$ diameter appears $1 \,cm$ from the surface nearest to the eye when looked at along the diameter. If the refractive index of glass is $_a\mu_g = 1.5$,the actual distance of the bubble from the refracting surface 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 $R$)

$A$ 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

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