To solve this problem, we need to understand the concept of dispersion without deviation.$\\$When two prisms are combined to produce dispersion without deviation, the net deviation is zero.$\\$For a thin prism, the deviation $\delta$ is given by:$\\ \delta = (\mu - 1)A \\$Where:$\\$• $\mu$ is the refractive index of the prism material.$\\$• $A$ is the refracting angle of the prism.$\\$Given:$\\$• First prism: $A_1 = 10^\circ, \mu_1 = 1.42 \\$• Second prism: $\mu_2 = 1.7 \\$Let the refracting angle of the second prism be $A_2. \\$For dispersion without deviation, the net deviation is zero:$\\ \delta_1 + \delta_2 = 0 \\$Substitute the deviation expressions for each prism:$\\ (\mu_1 - 1)A_1 + (\mu_2 - 1)A_2 = 0 \\$Substitute the given values:$\\ (1.42 - 1) \cdot 10^\circ + (1.7 - 1) \cdot A_2 = 0 \\ 0.42 \cdot 10^\circ + 0.7 \cdot A_2 = 0 \\ 4.2^\circ + 0.7A_2 = 0 \\$Rearrange to solve for $A_2: \\ 0.7A_2 = -4.2^\circ \\ A_2 = \frac{-4.2^\circ}{0.7} \\ A_2 = -6^\circ \\$Since the angle cannot be negative, we consider the magnitude:$\\ A_2 = 6^\circ \\$Therefore, the refracting angle of the second prism should be $6^\circ. \\$This corresponds to Option 1.