$A$ point object is placed at the centre of a glass sphere of radius $6 \, cm$ and refractive index $1.5$. The distance of the virtual image from the surface of the sphere is.....$cm$

Two identical glass rods $S_1$ and $S_2$ (refractive index $= 1.5$) have one convex end of radius of curvature $10 \ cm$. They are placed with the curved surfaces at a distance $d$ as shown in the figure,with their axes (shown by the dashed line) aligned. When a point source of light $P$ is placed inside rod $S_1$ on its axis at a distance of $50 \ cm$ from the curved face,the light rays emanating from it are found to be parallel to the axis inside $S_2$. The distance $d$ is (in $cm$)

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:
$(1) H_3 > H_1$
$(2) 0.8 \text{ cm} < (H_2 - H_1) < 0.9 \text{ cm}$
$(3) H_2 > H_3$
$(4) H_2 > H_1$

Two refracting media are separated by a spherical interface as shown in the figure. $PP'$ is the principal axis, $\mu_1$ and $\mu_2$ are the refractive indices of the medium of incidence and the medium of refraction respectively. Then:

$A$ hemispherical glass lens with a refractive index of $1.5$ is placed in a liquid with a refractive index of $1.3$ (see the figure). The radius of the hemispherical lens is $10 \,cm$. $A$ parallel beam of light traveling in the liquid is refracted by the glass lens. The absolute value of the position of the image from the center of the glass lens will be (in $\,cm$)

$A$ beam of diameter $d$ is incident on a glass hemisphere as shown. If the radius of curvature of the hemisphere is very large in comparison to $d$,then the diameter of the beam at the base of the hemisphere will be: