**Answer:**

The ratio of intensities of two waves causing interference is given as 9:4. Let's assume the maximum intensity is 9x and the minimum intensity is 4x.

Interference of two waves occurs when two wavefronts coincide. At the points of constructive interference, the amplitudes of the two waves add up, resulting in maximum intensity. At the points of destructive interference, the amplitudes of the two waves cancel each other out, resulting in minimum intensity.

Now, let's consider the superposition of the two waves at a point of constructive interference:

- At a point of constructive interference, the crest of one wave coincides with the crest of the other wave, resulting in maximum amplitude.
- The maximum amplitude occurs when the two waves are in phase, meaning their crests and troughs align perfectly.
- The amplitude of the resultant wave is the sum of the amplitudes of the two interfering waves.

Similarly, let's consider the superposition of the two waves at a point of destructive interference:

- At a point of destructive interference, the crest of one wave coincides with the trough of the other wave, resulting in minimum amplitude.
- The minimum amplitude occurs when the two waves are completely out of phase, meaning their crests align with the troughs.
- The amplitude of the resultant wave is the difference between the amplitudes of the two interfering waves.

**Calculating the ratios:**
- For constructive interference, the maximum intensity is given by the square of the sum of the amplitudes, which is (9x + 4x)^2 = 169x^2.
- For destructive interference, the minimum intensity is given by the square of the difference of the amplitudes, which is (9x - 4x)^2 = 25x^2.

Therefore, the ratio of the maximum and minimum intensities is (169x^2)/(25x^2) = 169/25 = 25:1.

Hence, the correct answer is option D) 25:1.

I α a^2 ==>I1/I2 =(a1/a2)^2 -->a1/a2 =3/2,Imax /Imin =[(a1+a2)/(a1-a2)]^2 =[(3+2)/(3-2)]^2 =25:1