A

Step 1: In Young's double-slit experiment, the intensity at point O can be described using the formula for intensity when there is a film in front of one of the slits. The intensity is affected by the refractive index of the film (n = 1.5) and the thickness (t) of the film.

Step 2: The intensity at point O initially is I0 and after placing the film, it becomes 0.75I0.
The intensity formula with a thin film is given by:
I = I_0 cos^2(\frac{\pi d}{\lambda})
where d is the path difference introduced by the film.

Step 3: The path difference d for the film is given by:
d = 2nt,
where n is the refractive index and t is the thickness of the film. Substituting n = 1.5, we get:
d = 3t.

Step 4: Since the intensity reduces to 75% of its original, we can set up the equation from the intensity formula:
0.75 I_0 = I_0 cos^2(\frac{3\pi t}{\lambda}).

Dividing both sides by I_0 gives us:
0.75 = cos^2(\frac{3\pi t}{\lambda}).

Step 5: Taking the square root gives us:
\sqrt{0.75} = cos(\frac{3\pi t}{\lambda}).
The value of \sqrt{0.75} = \frac{\sqrt{3}}{2}.

Step 6: From trigonometric values, the argument for cos equals to \frac{3\pi t}{\lambda} gives us possible angles. Let's consider cos inverse for simplicity and consider \frac{3\pi t}{\lambda} = \frac{\pi}{6}.

Step 7: Rearranging gives us:
t = \frac{\lambda}{18}.
Substituting \lambda = 720 nm, we get:
t = \frac{720 nm}{18} = 40 nm.

Therefore, the minimum thickness of the film is 40 nm. Thus, the correct answer is option A.