One of the refractive surfaces of a prism of angle $30^{\circ}$ is silvered. $A$ ray of light incident at an angle of $60^{\circ}$ retraces its path. The refractive index of the material of the prism is:

Write the equation for the refractive index of a prism in terms of the angle of minimum deviation $(\delta_m)$ and the angle of the prism $(A)$.

$A$ parallel beam of light is incident on a thin prism with a prism angle of $\frac{4}{\pi}$ degrees. The refractive index of the prism is $1.5$. The focal length of the lens is $60 \ cm$. The coordinates of the converging point of the beam are:

Light rays from a source are incident on a glass prism of refractive index $\mu$ and prism angle $\alpha$. At near-normal incidence,the angle of deviation of the emerging rays is:

$A$ prism of refractive index $n_{1}$ and another prism of refractive index $n_{2}$ are stuck together (as shown in the figure). $n_{1}$ and $n_{2}$ depend on $\lambda$, the wavelength of light, according to the relation:
${n}_{1}=1.2+\frac{10.8 \times 10^{-14}}{\lambda^{2}} \text{ and } {n}_{2}=1.45+\frac{1.8 \times 10^{-14}}{\lambda^{2}}$
The wavelength for which rays incident at any angle on the interface $BC$ pass through without bending at that interface will be $....\,nm.$

If the refractive index of the material of a prism is $\cot(A/2)$,where $A$ is the angle of the prism,then the angle of minimum deviation will be: