If the velocity of light in the first medium is $v_1$ and in the second medium is $v_2$,then the refractive index of the second medium with respect to the first medium is given by:

Shown in the figure is a transparent tank of length $30 \,cm$. $A$ black strip of width $3.8 \,cm$ is stuck on its left wall. When a source of light is kept to the left of it,a shadow of width $7.6 \,cm$ is formed on the right wall. Now,the tank is filled with a liquid of refractive index $n$,and the width of the shadow reduces to $6.4 \,cm$. The value of $n$ is closest to

Electromagnetic radiation of frequency $n$,wavelength $\lambda$,travelling with velocity $v$ in air,enters a glass slab of refractive index $\mu$. The frequency,wavelength,and velocity of light in the glass slab will be respectively:

The refractive index of a medium is $\frac{3}{2}$. The speed of light in this medium is . . . . . . $ms^{-1}$. (Speed of light in vacuum is $c = 3 \times 10^{8} \,ms^{-1}$)

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:

The optical path of a monochromatic light is the same if it goes through $4.0 \, cm$ of glass or $4.5 \, cm$ of water. If the refractive index of glass is $1.53$,the refractive index of water is: