$A$ red colour in air has a wavelength of $760 \,nm$. When light passes through water of refractive index $n = \frac{4}{3}$,its wavelength becomes $570 \,nm$. (The wavelength of yellow colour in air is $570 \,nm$). The colour of the red light in water is:

$A$ ray of light passes from the vacuum into a medium of refractive index $n$. If the angle of incidence is twice the angle of refraction, then the angle of incidence in terms of refractive index is

Find the refractive index of glass from the given figure,where the refractive index of water is $\mu_w = 4/3$.

If $\varepsilon_{0}$ and $\mu_{0}$ are the permittivity and permeability of free space and $\varepsilon$ and $\mu$ are the corresponding quantities for a medium,then the refractive index of the medium is:

The refractive index of glass is $1.5$ and that of water is $1.3$. If the speed of light in water is $2.25 \times 10^8 \, m/s$,what is the speed of light in glass?

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: