Solution

Consider a proton moving at a small transverse distance $x$ from the beam axis.

Let the first plate be grounded and let the potential of the second plate be $V$. The electric field between the plates is therefore approximately

$$E=\frac{V}{d}.$$

Before entering the capacitor, the protons were accelerated through a potential difference $V_0$, so their initial kinetic energy is

$$K_0=eV_0.$$

Hence,

$$\frac{mv_0^2}{2}=eV_0.$$

We assume $V>0$, so a positive proton moving from the first plate toward the second one is decelerated.

According to the condition,

$$l\gg d.$$

Therefore, the focusing action is weak, and we will eventually be able to use the approximation

$$\frac{V}{V_0}\ll1.$$

For convenience, introduce

$$\varepsilon=\frac{V}{V_0}.$$

1. Deflection at the first slit

Far from the slit, the electric field is perpendicular to the capacitor plates. Near a narrow slit, however, the field lines bend, producing a small transverse component $E_{\perp}$.

Consider the fringe-field region near the first slit. For a proton at a transverse distance $x$ from the axis, the integral of the transverse field component along the longitudinal direction is of order

$$\int E_{\perp}\,dz=\frac{V}{d}x.$$

The change in transverse momentum is

$$\Delta p_{\perp}=\int F_{\perp}\,dt.$$

Since

$$F_{\perp}=eE_{\perp},$$

we have

$$\Delta p_{\perp}=e\int E_{\perp}\,dt.$$

The fringe-field region is narrow, so the longitudinal speed may be treated as approximately constant while the proton crosses it:

$$dt=\frac{dz}{v}.$$

Therefore,

$$\Delta p_{\perp}=\frac{e}{v}\int E_{\perp}\,dz.$$

Using the field integral obtained above,

$$|\Delta p_{\perp}|=\frac{eVx}{dv}.$$

Let the initial transverse distance from the beam axis be $x_0$. At the first slit the proton speed is $v_0$, so

$$p_{\perp1}=\frac{eVx_0}{dv_0}.$$

Since the deflection angle is small,

$$\theta_1\simeq\frac{p_{\perp1}}{mv_0}.$$

Hence,

$$\theta_1=\frac{eVx_0}{dmv_0^2}.$$

Using

$$mv_0^2=2eV_0,$$

we obtain

$$\theta_1=\frac{Vx_0}{2V_0d}.$$

Thus,

$$\boxed{\theta_1=\frac{\varepsilon x_0}{2d}}.$$

The first slit slightly deflects the proton away from the axis.

2. Motion between the plates

Between the plates, the electric field is approximately uniform and directed along the longitudinal direction. Therefore, there is no transverse force, and the transverse momentum remains constant:

$$p_{\perp}=\text{const}.$$

However, the longitudinal speed decreases because the positive proton moves toward a higher electric potential.

At a distance $z$ from the first plate, the potential is

$$\varphi(z)=\frac{V}{d}z.$$

Energy conservation gives

$$\frac{mv^2(z)}{2}+e\varphi(z)=eV_0.$$

Therefore,

$$\frac{mv^2(z)}{2}=eV_0-e\frac{V}{d}z.$$

Factoring out $eV_0$,

$$\frac{mv^2(z)}{2}=eV_0\left(1-\varepsilon\frac{z}{d}\right).$$

Hence,

$$v(z)=v_0\sqrt{1-\varepsilon\frac{z}{d}}.$$

The transverse momentum remains unchanged after the first slit, so the transverse velocity is

$$v_{\perp}=\theta_1v_0.$$

Therefore, the slope of the trajectory inside the capacitor is

$$\frac{dx}{dz}=\frac{v_{\perp}}{v(z)}.$$

Substituting the expressions for $v_{\perp}$ and $v(z)$,

$$\frac{dx}{dz}=\frac{\theta_1}{\sqrt{1-\varepsilon z/d}}.$$

The transverse coordinate just before the second slit is therefore determined by

$$x_2-x_0=\int_0^d\frac{\theta_1\,dz}{\sqrt{1-\varepsilon z/d}}.$$

Taking $\theta_1$ outside the integral,

$$x_2-x_0=\theta_1\int_0^d\frac{dz}{\sqrt{1-\varepsilon z/d}}.$$

The integral is

$$\int_0^d\frac{dz}{\sqrt{1-\varepsilon z/d}}=\frac{2d}{\varepsilon}\left(1-\sqrt{1-\varepsilon}\right).$$

Thus,

$$x_2-x_0=\theta_1\frac{2d}{\varepsilon}\left(1-\sqrt{1-\varepsilon}\right).$$

Using

$$\theta_1=\frac{\varepsilon x_0}{2d},$$

we get

$$x_2-x_0=x_0\left(1-\sqrt{1-\varepsilon}\right).$$

Therefore,

$$\boxed{x_2=x_0\left(2-\sqrt{1-\varepsilon}\right)}.$$

3. Trajectory angle before the second slit

At the second plate, the proton kinetic energy is

$$K_2=e(V_0-V).$$

Therefore,

$$\frac{mv_2^2}{2}=e(V_0-V).$$

Using $\varepsilon=V/V_0$,

$$v_2=v_0\sqrt{1-\varepsilon}.$$

Since the transverse momentum remains constant between the plates,

$$\theta_{2-}=\frac{p_{\perp1}}{mv_2}.$$

But

$$\frac{p_{\perp1}}{mv_0}=\theta_1.$$

Hence,

$$\theta_{2-}=\theta_1\frac{v_0}{v_2}.$$

Therefore,

$$\boxed{\theta_{2-}=\frac{\varepsilon x_0}{2d\sqrt{1-\varepsilon}}}.$$

4. Deflection at the second slit

The second slit deflects the proton in the opposite direction, toward the beam axis.

Using the same expression for the transverse momentum change,

$$|\Delta p_{\perp2}|=\frac{eVx_2}{dv_2}.$$

The corresponding angular change is

$$|\Delta\theta_2|=\frac{|\Delta p_{\perp2}|}{mv_2}.$$

Thus,

$$|\Delta\theta_2|=\frac{eVx_2}{dmv_2^2}.$$

Since

$$mv_2^2=2e(V_0-V),$$

we obtain

$$|\Delta\theta_2|=\frac{Vx_2}{2d(V_0-V)}.$$

In terms of $\varepsilon$,

$$\boxed{|\Delta\theta_2|=\frac{\varepsilon x_2}{2d(1-\varepsilon)}}.$$

Because the second slit deflects the proton toward the axis,

$$\theta_{\text{out}}=\theta_{2-}-|\Delta\theta_2|.$$

Substituting the expressions obtained above,

$$\theta_{\text{out}}=\frac{\varepsilon x_0}{2d\sqrt{1-\varepsilon}}-\frac{\varepsilon x_2}{2d(1-\varepsilon)}.$$

Using

$$x_2=x_0\left(2-\sqrt{1-\varepsilon}\right),$$

we find

$$\theta_{\text{out}}=\frac{\varepsilon x_0}{2d}\left[\frac{1}{\sqrt{1-\varepsilon}}-\frac{2-\sqrt{1-\varepsilon}}{1-\varepsilon}\right].$$

Let

$$s=\sqrt{1-\varepsilon}.$$

Then

$$1-\varepsilon=s^2.$$

The expression in brackets becomes

$$\frac{1}{s}-\frac{2-s}{s^2}.$$

Combining the terms,

$$\frac{1}{s}-\frac{2-s}{s^2}=\frac{s-(2-s)}{s^2}=\frac{2s-2}{s^2}.$$

Therefore,

$$\theta_{\text{out}}=\frac{\varepsilon x_0}{d}\frac{s-1}{s^2}.$$

Returning to $\varepsilon$,

$$\boxed{\theta_{\text{out}}=-\frac{\varepsilon x_0}{d}\frac{1-\sqrt{1-\varepsilon}}{1-\varepsilon}}.$$

The minus sign shows that after the second slit the proton travels toward the beam axis.

5. Using the condition $l\gg d$

Since

$$l\gg d,$$

the system acts as a weak electrostatic lens, so

$$\varepsilon=\frac{V}{V_0}\ll1.$$

Expand the square root:

$$\sqrt{1-\varepsilon}=1-\frac{\varepsilon}{2}-\frac{\varepsilon^2}{8}+\ldots$$

Hence,

$$1-\sqrt{1-\varepsilon}=\frac{\varepsilon}{2}+\frac{\varepsilon^2}{8}+\ldots$$

Since the expression for $\theta_{\text{out}}$ already contains one factor of $\varepsilon$, it is sufficient to keep only

$$1-\sqrt{1-\varepsilon}\simeq\frac{\varepsilon}{2},$$

and also

$$1-\varepsilon\simeq1.$$

Therefore,

$$\theta_{\text{out}}\simeq-\frac{\varepsilon x_0}{d}\frac{\varepsilon}{2}.$$

Thus,

$$\boxed{\theta_{\text{out}}\simeq-\frac{\varepsilon^2x_0}{2d}}.$$

In magnitude,

$$\boxed{|\theta_{\text{out}}|\simeq\frac{x_0}{2d}\left(\frac{V}{V_0}\right)^2}.$$

Thus, the first-order effects of the two slits nearly cancel, and the resulting focusing effect is of second order in $V/V_0$.

6. Focusing condition

Beyond the second plate, the electric field is negligible, so the proton travels in a straight line.

If it reaches the beam axis at a distance $l$ from the second plate, then for a small angle,

$$|\theta_{\text{out}}|\simeq\frac{x_2}{l}.$$

Since

$$x_2=x_0\left[1+O(\varepsilon)\right],$$

while the angle itself is already of order $\varepsilon^2$, we may write, to the required accuracy,

$$x_2\simeq x_0.$$

Therefore,

$$|\theta_{\text{out}}|\simeq\frac{x_0}{l}.$$

On the other hand,

$$|\theta_{\text{out}}|\simeq\frac{x_0}{2d}\left(\frac{V}{V_0}\right)^2.$$

Equating these expressions,

$$\frac{x_0}{l}=\frac{x_0}{2d}\left(\frac{V}{V_0}\right)^2.$$

Canceling $x_0$,

$$\frac{1}{l}=\frac{1}{2d}\left(\frac{V}{V_0}\right)^2.$$

Hence,

$$\left(\frac{V}{V_0}\right)^2=\frac{2d}{l}.$$

Therefore,

$$\boxed{V=V_0\sqrt{\frac{2d}{l}}}.$$