Interference of Waves:

When two waves of similar frequency and amplitude meet, they superimpose each other and produce interference. The resulting amplitude at a point is the sum of the amplitudes of the two waves. The interference may be constructive or destructive depending on the phase difference between the two waves.

Intensity of Waves:

The intensity of a wave is the power per unit area. It is defined as the amount of energy passing through a unit area perpendicular to the direction of propagation of the wave in unit time. Intensity is proportional to the square of the amplitude of the wave.

Ratio of Intensities:

Given that the ratio of intensities of two waves causing interference is 9:4.

Let I1 and I2 be the intensities of the two waves.

Therefore, I1/I2 = 9/4

We can write I1 = (9/4)I2

Ratio of Maximum and Minimum Intensities:

The maximum intensity occurs when the two waves are in phase, and the minimum intensity occurs when they are out of phase. The amplitude of the resultant wave varies between the sum and difference of the amplitudes of the two waves.

Let A be the amplitude of the first wave and B be the amplitude of the second wave.

When the waves are in phase, the amplitude of the resultant wave is A+B.

When the waves are out of phase, the amplitude of the resultant wave is A-B.

The maximum intensity is proportional to (A+B)^2, and the minimum intensity is proportional to (A-B)^2.

Therefore, the ratio of maximum and minimum intensities is proportional to [(A+B)^2]/[(A-B)^2].

Substituting the values of A and B in terms of I1 and I2, we get

(A+B)^2 = (4I1+2I2)^2

(A-B)^2 = (4I1-2I2)^2

Ratio of maximum and minimum intensities = [(4I1+2I2)^2]/[(4I1-2I2)^2]

= [(2(2I1+I2))^2]/[(2(2I1-I2))^2]

= [(2I1+I2)/(2I1-I2)]^2

= [(9/4+1)/(9/4-1)]^2

= (13/5)^2

= 169/25

= 25/1

Therefore, the required ratio of maximum and minimum intensities is 25:1.

Hence, option D is the correct answer.