Refractive index of a medium is $\mu$. If the angle of incidence is twice that of the angle of refraction,then the angle of incidence is

Light enters from air into a given medium at an angle of $45^{\circ}$ with the interface of the air-medium surface. After refraction,the light ray is deviated through an angle of $15^{\circ}$ from its original direction. The refractive index of the medium is

Most materials have a refractive index,$n > 1$. So,when a light ray from air enters a naturally occurring material,then by Snell's law,$\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}$,it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism,the refractive index of the medium is given by the relation,$n = \left(\frac{c}{v}\right) = \pm \sqrt{\varepsilon_r \mu_r}$. Where $\varepsilon_r$ and $\mu_r$ are negative,one must choose the negative root of $n$. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behavior,without violating any physical laws. Since $n$ is negative,it results in a change in the direction of propagation of the refracted light. However,similar to normal materials,the frequency of light remains unchanged upon refraction even in meta-materials.
$1.$ Choose the correct statement.
$(A)$ The speed of light in the meta-material is $v = c|n|$.
$(B)$ The speed of light in the meta-material is $v = \frac{c}{|n|}$.
$(C)$ The speed of light in the meta-material is $v = c$.
$(D)$ The wavelength of the light in the meta-material $(\lambda_m)$ is given by $\lambda_m = \frac{\lambda_{\text{air}}}{|n|}$,where $\lambda_{\text{air}}$ is the wavelength of the light in air.
$2.$ For light incident from air on a meta-material,the appropriate ray diagram is:

An observer can see the top of a thin rod of height $h$ through a pinhole. The height of the container is $3h$ and its radius is $2h$. When the container is filled with a liquid up to a height of $2h$,the observer can see the bottom end of the rod. Find the refractive index of the liquid.

Captain Jack Sparrow tries to observe a fish almost vertically below him in a magical sea of variable refractive index $\mu = y^2 + 1$,where $y$ is the depth below the water surface. Find the apparent depth of the fish below the water level as seen by Captain Jack Sparrow. The actual depth of the fish is $1 \ m$.

The phase difference between the incident wave and the reflected wave is $180^o$ when a light ray: