If both the prisms are at minimum deviation condition, then $\dfrac{n_2}{n_1} = \dfrac{\sin\left(\dfrac{A_1}{2}\right)}{\sin\left(\dfrac{A_2}{2}\right)}$.

If prism 2 is at minimum deviation condition, then $\sin i_1 = n_2 \sin \left(\dfrac{A_2}{2}\right)$ is always true.

If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation $\delta_{m1}$ and $\delta_{m2}$, respectively, then $\theta = \dfrac{\delta_{m1}}{2(n_1 - 1)} + \dfrac{\delta_{m2}}{2(n_2 - 1)}$.

If prism 1 is at minimum deviation condition, then $\sin i_2 = n_1 \sin \left(\dfrac{A_1}{2}\right)$ is always true.

The vertical distance of the particle from the $XZ$ plane at $t = 0.3\ s$ is $15\ \mathrm{cm}$.

The vertical distance of the particle from the $XZ$ plane at $t = 0.4\ s$ is $10\ \mathrm{cm}$.

The radius of the trajectory of the particle for $t > 0.2\ s$ is $20\ \mathrm{cm}$.

The particle will be in the $XZ$ plane at $t = 0.35\ s$.

For $m = -1$, locus of these points is $ax + by = 0$.

For $m = 2$, the locus of these points is a circle of radius $\frac{2}{3}\sqrt{a^2+b^2}$ centered at $\left(\frac{2}{3}a, \frac{2}{3}b\right)$

For $m = -2$, the locus of these points is a circle of radius $2\sqrt{a^2+b^2}$ centered at $(2a, 2b)$

For $m = -3$, locus of these points is $3bx - 3ay = 0$.

The center of mass of the dipole is deflected towards $\hat{\jmath}$ in the presence of the field.

If the magnitude of the final angular velocity $\omega_f = \sqrt{\frac{2qE}{md}}$, then $\theta_f = \frac{\pi}{6}$.

If $\theta_f = \pi/3$, then the change in kinetic energy of the dipole is given by $2\sqrt{3}\ qEd$.

For $\theta_f = \pi/4$, the dipole rotates around its center of mass with a constant angular velocity after $t > t_f$.