The angle of minimum deviation measured with a prism is $30^o$ and the angle of the prism is $60^o$. The refractive index of the prism material is:

For a thin prism,$\delta_1$ is the angle of deviation produced when the prism is placed in air. When the prism is immersed in water,the angle of deviation produced is $\delta_2$. Given ${ }_{a} \mu_{g}=\frac{3}{2}$ and ${ }_{a} \mu_{w}=\frac{4}{3}$. The ratio $\delta_2: \delta_1$ is

The prism angle of a prism is $40^{\circ}$. $A$ light ray is incident at $38^{\circ}$ and undergoes minimum deviation. The angle of minimum deviation is ........$^{\circ}$.

$A$ ray of light passes through an equilateral prism such that the angle of incidence $(i)$ is equal to the angle of emergence $(e)$. The angle of emergence is equal to $\left(\frac{3}{4}\right)$ of the angle of the prism. The angle of deviation is: (in $^{\circ}$)

The ratio of the angle of minimum deviation of a prism in air to that when it is dipped in water is (given $_a\mu_g = 3/2$ and $_a\mu_w = 4/3$).

Three prisms $1, 2$ and $3$ have the prism angle $A = 60^o$,but their refractive indices are respectively $1.4, 1.5$ and $1.6$. If $\delta_1, \delta_2, \delta_3$ be their respective angles of deviation,then: