A photosensitive material is at 9m to the left of the origin and the source of light is at 7 m to the right of the origin along x-axis.The photosensitive material and the source of light start from rest and move, respectively, with `8hatims^(-1)` and `4hatims^(-1)`. The ratio of intensities at `t=0` to `t-3 s` as received by the photosensitive material

Options

A. `16:1`
B. `1:16`
C. `2:7`
D. `7:2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial positions
- The photosensitive material is at -9 m (9 m to the left of the origin).
- The source of light is at +7 m (7 m to the right of the origin).

### Step 2: Calculate the initial distance between the source and the photosensitive material at t = 0
- The initial distance \( d_1 \) is given by:
\[ d_1 = |(-9) - (7)| = |-9 - 7| = | -16 | = 16 \text{ m} \]

### Step 3: Determine the positions of the source and photosensitive material at t = 3 s
- The source moves with a velocity of \( 4 \, \text{m/s} \):
\[ \text{Position of source at } t = 3 \text{ s} = 7 + (4 \times 3) = 7 + 12 = 19 \text{ m} \]
- The photosensitive material moves with a velocity of \( 8 \, \text{m/s} \):
\[ \text{Position of photosensitive material at } t = 3 \text{ s} = -9 + (8 \times 3) = -9 + 24 = 15 \text{ m} \]

### Step 4: Calculate the distance between the source and the photosensitive material at t = 3 s
- The new distance \( d_2 \) is given by:
\[ d_2 = |15 - 19| = |-4| = 4 \text{ m} \]

### Step 5: Use the inverse square law to find the ratio of intensities
- The intensity ratio \( \frac{I_1}{I_2} \) can be expressed as:
\[ \frac{I_1}{I_2} = \frac{d_2^2}{d_1^2} \]
- Substitute the values:
\[ \frac{I_1}{I_2} = \frac{4^2}{16^2} = \frac{16}{256} = \frac{1}{16} \]

### Conclusion
- The ratio of intensities at \( t = 0 \) to \( t = 3 \) seconds is \( \frac{1}{16} \).