The most appropriate reason for the use of prism in preference to a mirror in a periscope is that:

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $n_1$ surrounded by a medium of lower refractive index $n_2$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $n_1$ and $n_2$ as shown in the figure. All rays with the angle of incidence $i$ less than a particular value $i_m$ are confined in the medium of refractive index $n_1$ . The numerical aperture (NA) of the structure is defined as $\sin i_m$.

If two structures of same cross-sectional area, but different numerical apertures $\mathrm{N}{\mathrm{A}}_1$ and $\mathrm{N}{\mathrm{A}}_2$ $\left(\mathrm{N}{\mathrm{A}}_2<\mathrm{N}{\mathrm{A}}_1\right)$ are joined longitudinally, the numerical aperture of the combined structure is

In an optical fibre, the propagation of light occurs due to which phenomenon? Write answer only.

The principle used in the transmission of signals through an optical fibre is

Three rays of light, namely red $\left(R\right)$, green $\left(G\right)$ and blue $\left(B\right)$ are incident on the face $PQ$ of a right angled prism $PQR$ as shown in figure.

The refractive indices of the material of the prism for red, green and blue wavelength are $1.27,1.42$ and $1.49$ respectively. The colour of the ray(s) emerging out of the face $PR$ is :

For a right angled prism ray $1$is the incident ray and ray $2$ is the emergent ray, as shown in the figure. The refractive index of the prism is ____.

What is optical fibre? Write its two important uses. 

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $n_1$ surrounded by a medium of lower refractive index $n_2$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $n_1$ and $n_2$ as shown in the figure. All rays with the angle of incidence $i$ less than a particular value $i_m$ are confined in the medium of refractive index $n_1$ . The numerical aperture (NA) of the structure is defined as $\sin i_m$.

For two structures namely $S_1$ with $n_1=\frac{\sqrt{45}}{4}$ and $n_2=\frac{3}{2}$ , and $S_2$ with $n_1=\frac{8}{5}$ and $n_2=\frac{7}{5}$ and taking the refractive index of water to be $\frac{4}{3}$ and that of air to be 1, the correct option(s) is (are)