MODERN BYU 5 - CHATUP707

Solution: Neutron Beam Interference in Tilted Square

Solution for Question 5

1. Phase Difference Analysis

The neutron beam splits at point O into two paths: $O \to A \to C$ and $O \to B \to C$. Due to the tilt angle $\theta$, the upper path BC is at a higher gravitational potential than the lower path OA.

The momentum of the neutron depends on its kinetic energy $K = E_{total} – U_{grav}$. In the inclined sections OB and AC, the neutron experiences the same change in vertical height, so the phase accumulated in these two segments is identical. Therefore, the net phase difference between the two paths arises solely from the difference in momentum between the bottom horizontal side OA and the top horizontal side BC.

The condition for constructive interference (a maximum) at point C is that the path difference measured in wavelengths must be an integer $n$:

$$ n = \frac{l}{\lambda_{OA}} – \frac{l}{\lambda_{BC}} $$

2. Momentum Relations

Using the de Broglie relationship $\lambda = h/p$, we can rewrite the interference condition in terms of momentum:

$$ n = \frac{l}{h/p_{OA}} – \frac{l}{h/p_{BC}} = \frac{l}{h} (p_{OA} – p_{BC}) $$ $$ \frac{nh}{l} = p_{OA} – p_{BC} $$

We express the momentum $p$ in terms of the total energy $E_0$ and gravitational potential energy. $p = \sqrt{2mK} = \sqrt{2m(E_{total} – U)}$.

At the bottom level (path OA), the potential energy is zero (by defining O at height 0). So, $p_{OA} = \sqrt{2mE_0}$.
At the top level (path BC), the height is $h’ = l \sin\theta$. The potential energy is $U = mgh’ = mgl\sin\theta$. The kinetic energy is $K_{BC} = E_0 – mgl\sin\theta$. Thus, $p_{BC} = \sqrt{2m(E_0 – mgl\sin\theta)}$.

3. Algebraic Solution

Substituting the expressions for $p_{OA}$ and $p_{BC}$ into our interference condition:

$$ \frac{nh}{l} = \sqrt{2mE_0} – \sqrt{2m(E_0 – mgl\sin\theta)} $$

Rearranging to isolate the term with $\sin\theta$:

$$ \sqrt{2m(E_0 – mgl\sin\theta)} = \sqrt{2mE_0} – \frac{nh}{l} $$

Squaring both sides of the equation:

$$ 2m(E_0 – mgl\sin\theta) = \left( \sqrt{2mE_0} – \frac{nh}{l} \right)^2 $$ $$ 2mE_0 – 2m^2gl\sin\theta = 2mE_0 + \frac{n^2h^2}{l^2} – 2\sqrt{2mE_0} \left( \frac{nh}{l} \right) $$

The $2mE_0$ terms on both sides cancel out. We can then rearrange to solve for the term with $\sin\theta$:

$$ 2m^2gl\sin\theta = \frac{2nh}{l}\sqrt{2mE_0} – \frac{n^2h^2}{l^2} $$

4. Final Expression

To simplify the right-hand side, we find a common denominator of $l^2$:

$$ 2m^2gl\sin\theta = \frac{2nhl\sqrt{2mE_0} – n^2h^2}{l^2} $$

Finally, we divide by $2m^2gl$ to isolate $\sin\theta$ and then take the inverse sine to find $\theta$:

$$ \sin\theta = \frac{2nhl\sqrt{2mE_0} – n^2h^2}{2m^2gl^3} $$ $$ \theta = \sin^{-1} \left( \frac{2nhl\sqrt{2mE_0} – n^2 h^2}{2m^2gl^3} \right) $$

where $n = 1, 2, 3 \dots$ represents the order of the maxima.