Let the slit widths be proportional to amplitudes

Suppose the narrower slit has width $a$ and the wider slit has width $3a$.

Since amplitude is proportional to slit width, $A_1 : A_2 = 1 : 3$

Let $A_1 = A, \qquad A_2 = 3A$

Use interference intensity formulas

For two coherent sources with amplitudes $A_1$ and $A_2$:

• Maximum intensity occurs when waves interfere constructively: $I_{\max} = (A_1 + A_2)^2$

• Minimum intensity occurs when waves interfere destructively: $I_{\min} = (A_2 - A_1)^2$

Substitute the amplitudes

$I_{\max} = (A + 3A)^2 = (4A)^2 = 16A^2$

$I_{\min} = (3A - A)^2 = (2A)^2 = 4A^2$

Find the ratio

$I_{\min} : I_{\max} = 4A^2 : 16A^2 = 1 : 4$

This is given as $x : 4$.

Therefore, $x = 1$

Comparison with stored answer

Stored correct answer = $1$

Our derived answer also gives $x=1$. So they agree.