$A$ 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 $R$)

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.....$cm$.

In a medium of refractive index $1.6$ having a convex surface,a point object is placed at a distance of $12 \,cm$ from the pole. The radius of curvature of the surface is $6 \,cm$. Locate the image as seen from air.

Three glass cylinders of equal height $H = 30 \text{ cm}$ and same refractive index $n = 1.5$ are placed on a horizontal surface as shown in the figure. Cylinder $I$ has a flat top,cylinder $II$ has a convex top,and cylinder $III$ has a concave top. The radii of curvature of the two curved tops are same $(R = 3 \text{ m})$. If $H_1, H_2$ and $H_3$ are the apparent depths of a point $X$ on the bottom of the three cylinders,respectively,the correct statement$(s)$ is/are:
$(1) H_3 > H_1$
$(2) 0.8 \text{ cm} < (H_2 - H_1) < 0.9 \text{ cm}$
$(3) H_2 > H_3$
$(4) H_2 > H_1$

$A$ 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 . . . . . . $cm$.

$A$ point object is placed at a distance of $15 \, cm$ from a glass rod with a spherical end. The radius of curvature of the spherical end is $30 \, cm$. Find the position of the image formed.