Angle of deviation $(\delta)$ by a prism (refractive index = $\mu$ and supposing the angle of prism $A$ to be small) can be given by

For an isosceles prism of angle $A$ and refractive index $\mu$,it is found that the angle of minimum deviation $\delta_m=A$. Which of the following options is/are correct?
[$A$] At minimum deviation,the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1=\left(i_1 / 2\right)$.
[$B$] For this prism the refractive index $\mu$ and the angle of prism $A$ are related as $A=\frac{1}{2} \cos ^{-1}(\mu / 2)$.
[$C$] For this prism,the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1=\sin ^{-1}\left[\sin A \sqrt{4 \cos ^2 \frac{A}{2}-1}-\cos A\right]$.
[$D$] For the angle of incidence $i_1=A$,the ray inside the prism is parallel to the base of the prism.

In the visible region, the dispersive powers and the mean angular deviations for crown and flint glass prisms are $\omega, \omega^{\prime}$ and $d, d^{\prime}$ respectively. The condition for obtaining dispersion without deviation, when the two prisms are combined, is:

The path of a refracted ray of light in a prism is parallel to the base of the prism only when the

Draw a diagram of a right-angled prism for which the object and image are of the same size.

The refractive index of a prism is $\mu = \sqrt{3}$ and the ratio of the angle of minimum deviation to the angle of the prism is $1$. The value of the angle of the prism is $......^{\circ}$.