$A$ ray of light incident on one face of an equilateral glass prism having refractive index $\sqrt{2}$, produces the emergent ray which just grazes along the adjacent face. The value of angle of incidence is $(\sin 90^{\circ} = 1, \sin 30^{\circ} = 0.5, \sin 45^{\circ} = 1/\sqrt{2})$.

The cross-section of a prism is an equilateral triangle $ABC$ as shown in the figure. The minimum deviation is observed when the angle of incidence is equal to the prism angle. The time taken by light to travel from the midpoint $P$ of $BC$ to $A$ is $..... \times 10^{-10} \, s$. (Given: speed of light in vacuum $= 3 \times 10^8 \, m/s$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$)

The angle of a prism is $60^{\circ}$. When light is incident at an angle of $60^{\circ}$ on the prism,the angle of emergence is $40^{\circ}$. The angle of incidence $i$ for which the light ray will deviate the least is such that

An achromatic prism is made by combining two prisms $P_1$ $(\mu_v = 1.523, \mu_r = 1.515)$ and $P_2$ $(\mu'_v = 1.666, \mu'_r = 1.650)$,where $\mu$ represents the refractive index. If the angle of the prism $P_1$ is $10^o$,then the angle of the prism $P_2$ will be.....$^o$

Assertion: There exist two angles of incidence for the same magnitude of deviation (except for minimum deviation) by a prism kept in air.
Reason: In a prism kept in air,a ray is incident on the first surface and emerges out of the second surface. If another ray is incident on the second surface along the path of the previous emergent ray,then this ray emerges out of the first surface along the path of the previous incident ray. This principle is called the principle of reversibility of light.

At what angle will a ray of light be incident on one face of an equilateral prism,so that the emergent ray may graze the second surface of the prism $(\mu=2)$ (in $^{\circ}$)?