A short straight object of height $100 \mathrm{cm}$ lies before the central axis of a spherical mirror whose focal length has absolute value $\left|f\right|=40 \mathrm{cm}$. The image of object produced by the mirror is of height $25 \mathrm{cm}$ and has the same orientation of the object. One may conclude from the information:

Given below are two statements : one is labeled as Assertion $A$ and the other is labeled as Reason $R$.

Assertion $A$: For a simple microscope, the angular size of the object equals the angular size of the image.

Reason $R$: Magnification is achieved as the small object can be kept much closer to the eye than $25 \mathrm{cm}$ and hence it subtends a large angle. In the light of the above statements, choose the most appropriate answer from the options given below:

A point source of light $S$, placed at a distance $60 \mathrm{cm}$ infront of the centre of a plane mirror of width $50 \mathrm{cm}$, hangs vertically on a wall. A man walks infront of the mirror along a line parallel to the mirror at a distance $1.2 \mathrm{m}$ from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is__$\mathrm{cm}$

A spherical mirror is obtained as shown in the figure from a hollow glass sphere, if an object is positioned in front of the mirror, what will be the nature and magnification of the image of the object? (Figure down as schematic and not to scale)

A light ray enters a solid glass sphere of refractive index $\mathrm{\mu }=\sqrt{3}$ at an angle of incidence $60°$. The ray is both reflected and refracted at the farther surface of the sphere. The angle (in degrees) between the reflected and refracted rays at this surface is:

An observer can see through a small hole on the side of a jar (radius $15 \mathrm{cm}$) at a point at height of $15 \mathrm{cm}$ from the bottom. The hole is at a height of $45 \mathrm{cm}$. When the jar is filled with a liquid up to a height of $30 \mathrm{cm}$ the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is $\sqrt{N}$, the value of $\mathrm{N}$ is: