The refractive indices of water and glass with respect to air are $1.2$ and $1.5$ respectively. The refractive index of glass with respect to water is

An opaque sphere of radius $a$ is just immersed in a transparent liquid as shown in the figure. $A$ point source is placed on the vertical diameter of the sphere at a distance $a/2$ from the top of the sphere. One ray originating from the point source after refraction from the air-liquid interface forms a tangent to the sphere. The angle of refraction for that particular ray is $30^{\circ}$. The refractive index of the liquid is:

In finding out the refractive index of a glass slab,the following observations were made through a travelling microscope: $50$ vernier scale divisions $= 49$ $MSD$; $20$ divisions on the main scale in each $cm$. For a mark on paper:
$MSR = 8.45 \ cm, VC = 26$
For the mark on paper seen through the slab:
$MSR = 7.12 \ cm, VC = 41$
For a powder particle on the top surface of the glass slab:
$MSR = 4.05 \ cm, VC = 1$
($MSR =$ Main Scale Reading,$VC =$ Vernier Coincidence)
The refractive index of the glass slab is:

If the ray $CD$ is parallel to the ray $AB$,then:

An observer can see through a small hole on the side of a jar (radius $15\, cm$) at a point at a height of $15\, cm$ from the bottom (see figure). The hole is at a height of $45\, cm$. When the jar is filled with a liquid up to a height of $30\, cm$,the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is $N/100$,where $N$ is an integer,the value of $N$ is $.....$

$A$ student sees the top edge and the bottom centre $C$ of a pool simultaneously from an angle $\theta$ above the horizontal as shown in the figure. The refractive index of water which fills up to the top edge of the pool is $\frac{4}{3}$. If $\frac{h}{x}=\frac{7}{4}$,then $\cos \theta$ is