$A$ 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 $9.4 \, cm$. The refractive index of water is .....

$A$ transparent slab of thickness $d$ has a refractive index $n(z)$ that increases with $z$. Here $z$ is the vertical distance inside the slab,measured from the top. The slab is placed between two media with uniform refractive indices $n_1$ and $n_2 (> n_1)$,as shown in the figure. $A$ ray of light is incident with angle $\theta_i$ from medium $1$ and emerges in medium $2$ with refraction angle $\theta_f$ with a lateral displacement $l$. Which of the following statement$(s)$ is(are) true?
$(A)$ $n_1 \sin \theta_i = n_2 \sin \theta_f$
$(B)$ $n_1 \sin \theta_i = (n_2 - n_1) \sin \theta_f$
$(C)$ $l$ is independent of $n_2$
$(D)$ $l$ is dependent on $n(z)$

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?

Time taken by light to travel in two different materials $A$ and $B$ of refractive indices $\mu_{A}$ and $\mu_{B}$ of the same thickness is $t_{1}$ and $t_{2}$ respectively. If $t_{2}-t_{1}=5 \times 10^{-10} \text{ s}$ and the ratio of $\mu_{A}$ to $\mu_{B}$ is $1:2$. Then the thickness of the material,in meters,is: (Given $v_{A}$ and $v_{B}$ are velocities of light in $A$ and $B$ materials respectively).

Light travels through water in a beaker. The height of the water column is $h$. If the refractive index of water is $\mu_{w}$ and $C$ is the velocity of light in air,the time taken by light to travel through the water will be: