Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} \text{ m/s}$ and $2.0 \times 10^{8} \text{ m/s}$ respectively. The critical angle between them is.

When a ray of light emerges from a block of glass,the critical angle is

$A$ right-angled isosceles prism is held on the surface of a liquid composed of miscible solvents $A$ and $B$ of refractive index $n_{A}=1.5$ and $n_{B}=1.3$,respectively. The refractive index of the prism is $n_{p}=1.5$ and that of the liquid is given by $n_{L}=C_{A} n_{A}+(1-C_{A}) n_{B}$,where $C_{A}$ is the percentage of solvent $A$ in the liquid. If $\theta_{C}$ is the critical angle at the prism-liquid interface,the plot which best represents the variation of the critical angle with the percentage of solvent is

$A$ wide slab consisting of two media of refractive indices $n_1$ and $n_2$ is placed in air as shown in the figure. $A$ ray of light is incident from medium $n_1$ to $n_2$ at an angle $\theta$,where $\sin \theta$ is slightly larger than $1/n_1$. Take the refractive index of air as $1$. Which of the following statement$(s)$ is(are) correct?
$(A)$ The light ray enters air if $n_2 = n_1$
$(B)$ The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 < n_1$
$(C)$ The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 > n_1$
$(D)$ The light ray is reflected back into the medium of refractive index $n_1$ if $n_2 = 1$

In the figure shown,for an angle of incidence $45^{\circ}$ at the top surface,what is the minimum refractive index $\mu$ needed for total internal reflection at the vertical face?

Assertion: Critical angle is minimum for violet colour.
Reason: Because critical angle $\theta_c = \sin^{-1} \left( \frac{1}{\mu} \right)$ and $\mu \propto \frac{1}{\lambda}$.