$A$ ray of laser of a wavelength $630 \, nm$ is incident at an angle of $30^{\circ}$ at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is $2.42$ and that of air is $1$. Choose the correct option.

  • A
    angle of refraction is $24.41^{\circ}$
  • B
    angle of refraction is $30^{\circ}$
  • C
    refraction is not possible
  • D
    angle of refraction is $53.4^{\circ}$

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The relation between the critical angles of water and glass is:

Light guidance in an optical fiber can be understood by considering a structure comprising of a 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 $n_2$. 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 $\sin i_m$.
$1.$ For two structures namely $S_1$ with $n_1=\sqrt{45}/4$ and $n_2=3/2$,and $S_2$ with $n_1=8/5$ and $n_2=7/5$,and taking the refractive index of water to be $4/3$ and that of air to be $1$,the correct option$(s)$ is(are):
$(A)$ $NA$ of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{16}{3\sqrt{15}}$
$(B)$ $NA$ of $S_1$ immersed in a liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_2$ immersed in water
$(C)$ $NA$ of $S_1$ placed in air is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{4}{\sqrt{15}}$
$(D)$ $NA$ of $S_1$ placed in air is the same as that of $S_2$ placed in water
$2.$ If two structures of the same cross-sectional area,but different numerical apertures $NA_1$ and $NA_2$ $(NA_2 < NA_1)$ are joined longitudinally,the numerical aperture of the combined structure is:
$(A)$ $\frac{NA_1 NA_2}{NA_1+NA_2}$ $(B)$ $NA_1+NA_2$ $(C)$ $NA_1$ $(D)$ $NA_2$

$A$ diver at a depth of $12 \ m$ in water $\left(\mu = \frac{4}{3}\right)$ sees the sky in a cone of semi-vertical angle:

On which principle does an optical fiber work?

'Mirage' is a phenomenon due to

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