1. Resolve the velocity into components

Let the electron enter the region between the parallel plates with speed $u$ and leave with speed $v$.

Since the electron enters at angle $\alpha$ with the plates, its component of velocity parallel to the plates is $u_\parallel = u\cos\alpha.$

Similarly, when it leaves at angle $\beta$ with the plates, the parallel component is $v_\parallel = v\cos\beta.$

2. Use the fact about motion in a capacitor

The electric field between parallel plates is perpendicular to the plates. Hence, there is no force parallel to the plates.

Therefore, the component of velocity parallel to the plates remains unchanged: $u\cos\alpha = v\cos\beta.$

So, $\frac{u}{v} = \frac{\cos\beta}{\cos\alpha}.$

3. Relate kinetic energies to speeds

Kinetic energy is $K = \frac12 m v^2.$

Thus, $K_1 = \frac12 m u^2, \qquad K_2 = \frac12 m v^2.$

Hence, $\frac{K_1}{K_2} = \frac{u^2}{v^2}.$

Using $\frac{u}{v} = \frac{\cos\beta}{\cos\alpha},$ we get $\frac{K_1}{K_2} = \left(\frac{\cos\beta}{\cos\alpha}\right)^2.$

Therefore, $K_1:K_2 = \frac{\cos^2\beta}{\cos^2\alpha}.$

4. Match with the options

This corresponds to:

A. $\displaystyle \frac{\cos^2\beta}{\cos^2\alpha}$

Answer: Option A