Light of wavelength $300 \ nm$ in medium $A$ enters into medium $B$ through a plane surface. If the frequency of light is $5 \times 10^{14} \ Hz$ and the ratio of speed in medium $A$ to that in medium $B$ is $\frac{4}{5}$,the absolute refractive index of medium $B$ is:

$A$ ray of light travelling in a medium of refractive index $\mu$ is incident at an angle $\theta$ on a composite transparent plate consisting of $50$ plates of refractive indices $1.01\mu, 1.02\mu, 1.03\mu, \dots, 1.50\mu$. The ray emerges from the composite plate into a medium of refractive index $1.6\mu$ at an angle $x$. Then:

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:

$A$ ray of light passes through four transparent media with refractive indices $\mu_1, \mu_2, \mu_3$ and $\mu_4$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $CD$ is parallel to the incident ray $AB$,we must have

On a glass plate,a light wave is incident at an angle of $60^o$. If the reflected and the refracted waves are mutually perpendicular,the refractive index of the material is:

When light enters glass from vacuum,then the wavelength of light: