When two prisms composed of different materials are properly...

When two prisms of different materials are combined, the net deviation is given by $\delta = \delta_1 + \delta_2 = A_1(\mu_1 - 1) + A_2(\mu_2 - 1)$ and the net dispersion is given by $\theta = \theta_1 + \theta_2 = A_1(\mu_{v1} - \mu_{r1}) + A_2(\mu_{v2} - \mu_{r2})$.

By appropriately choosing the angles $A_1$ and $A_2$ and placing the prisms in opposite orientations (one inverted with respect to the other), we can achieve the following conditions:

1. Deviation without dispersion (Achromatic combination): The condition for zero net dispersion is $\theta = 0$, which gives $A_1(\mu_{v1} - \mu_{r1}) = -A_2(\mu_{v2} - \mu_{r2})$. Since the materials are different, their dispersive powers $\omega_1$ and $\omega_2$ are different, so the net deviation $\delta$ will not be zero.

2. Dispersion without average deviation (Direct vision combination): The condition for zero net deviation is $\delta = 0$, which gives $A_1(\mu_1 - 1) = -A_2(\mu_2 - 1)$. Again, due to different dispersive powers, the net dispersion $\theta$ will not be zero.

3. Both dispersion and average deviation: If the prisms are combined without satisfying either of the above specific null conditions, the combination will generally produce both dispersion and deviation.

4. Neither dispersion nor average deviation: For both $\delta = 0$ and $\theta = 0$ to hold simultaneously, the dispersive powers of the two prisms must be equal ($\omega_1 = \omega_2$). Since the prisms are made of different materials, they have different dispersive powers, making this case impossible.

Therefore, it is possible to produce deviation without dispersion, dispersion without average deviation, and both dispersion and average deviation.