The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea $(7.8 \, mm)$. This surface separates two media of refractive indices $1$ and $1.34$. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus in $cm$.

$A$ ray coming from an object which is situated at infinity in the air falls on a spherical glass surface $(n=1.5)$. The distance of the image will be . . . . . . . ($R$ is the radius of curvature of the spherical glass surface.)

$A$ 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.....$cm$

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

$A$ 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 interface of radius $R$ separates two media of refractive indices $1$ and $1.4$ respectively, as shown in the figure below. $A$ point source is placed at a distance of $4R$ in front of the spherical interface. The magnitude of the lateral magnification of the point source image is . . . . . . .