$A$ pole is vertically submerged in a swimming pool such that it casts a shadow of length $2.15 \, m$ within the water when sunlight is incident at an angle of $30^{\circ}$ with the water surface. If the swimming pool is filled to a height of $1.5 \, m$,then the height of the pole above the water surface in centimeters is $(\mu_W = 4/3) .........$

$A$ beam of light is propagating at an angle $\alpha_1$ from medium $1$ into medium $2$ at an angle $\alpha_2$. If the wavelength of light in medium $1$ is $\lambda_1$, then the wavelength of light in medium $2$, $\lambda_2$, is:

Consider a light ray travelling in air incident into a medium of refractive index $\sqrt{2n}$. The incident angle is twice that of the refracting angle. Then,the angle of incidence will be

$A$ tank is filled with water to a height of $12.5 \;cm$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be $9.4 \;cm$. What is the refractive index of water?
If water is replaced by a liquid of refractive index $1.63$ up to the same height,by what distance would the microscope have to be moved to focus on the needle again?

$A$ ray of light is incident at the glass-water interface at an angle $i$. It emerges finally parallel to the surface of water. If the refractive index of water is $\mu_w = 4/3$,then the value of the refractive index of glass $\mu_g$ is:

$A$ ray of light of frequency $5 \times 10^{14} \,Hz$ is passed through a liquid. The wavelength of light measured inside the liquid is found to be $450 \times 10^{-9} \,m$. Calculate the refractive index of the liquid.