A long narrow horizontal slit is placed 1 mm above a horizontal...

Let the intensity of the light arriving directly from the slit be $I_1 = I$.

Since the mirror reflects $64\%$ of the incident light energy, the intensity of the light arriving after reflection is $I_2 = 0.64 I$.

The maximum intensity in the interference pattern is given by I_\max = (\sqrt{I_1} + \sqrt{I_2})^2.

The minimum intensity in the interference pattern is given by I_\min = (\sqrt{I_1} - \sqrt{I_2})^2.

The ratio of the maximum intensity to the minimum intensity is:

\dfrac{I_\max}{I_\min} = \left( \dfrac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2

Substituting the values of $I_1$ and $I_2$:

\dfrac{I_\max}{I_\min} = \left( \dfrac{\sqrt{I} + \sqrt{0.64 I}}{\sqrt{I} - \sqrt{0.64 I}} \right)^2

\dfrac{I_\max}{I_\min} = \left( \dfrac{1 + 0.8}{1 - 0.8} \right)^2

\dfrac{I_\max}{I_\min} = \left( \dfrac{1.8}{0.2} \right)^2 = (9)^2 = 81

Thus, the ratio is $81 : 1$.