Two monochromatic light beams of intensity 16 and 9 units are interfering. The ratio of intensities of bright and dark parts of the resultant pattern is:

Two monochromatic light beams with intensities $$I_1 = 16$$ units and $$I_2 = 9$$ units interfere. We need to find the ratio of the intensity of the bright parts to the intensity of the dark parts in the resultant interference pattern.

For interference of two coherent monochromatic light beams, the intensity at any point is given by:

$$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$$

where $$\phi$$ is the phase difference between the two beams.

The maximum intensity (bright parts) occurs when $$\cos \phi = 1$$, so:

$$I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2}$$

The minimum intensity (dark parts) occurs when $$\cos \phi = -1$$, so:

$$I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2}$$

We are asked for the ratio $$ \frac{I_{\text{max}}}{I_{\text{min}}} $$.

First, calculate $$\sqrt{I_1 I_2}$$:

$$I_1 = 16, \quad I_2 = 9$$

$$\sqrt{I_1 I_2} = \sqrt{16 \times 9} = \sqrt{144} = 12$$

Now substitute into the formula for $$I_{\text{max}}$$:

$$I_{\text{max}} = 16 + 9 + 2 \times 12 = 25 + 24 = 49$$

Next, substitute into the formula for $$I_{\text{min}}$$:

$$I_{\text{min}} = 16 + 9 - 2 \times 12 = 25 - 24 = 1$$

Therefore, the ratio is:

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{49}{1} = 49$$

Now, comparing with the options:

A. $$\frac{16}{9}$$

B. $$\frac{4}{3}$$

C. $$\frac{7}{1}$$

D. $$\frac{49}{1}$$

The ratio $$\frac{49}{1}$$ matches option D.

Hence, the correct answer is Option D.