Three coherent, equal intensity light rays arrive at a point Pon a screen to produce an interference minimum of zero intensity. If any two of the rays are blocked, the intensity of the light at P is `I_(1)`. What is the intensity of the light at P if only one of the rays is blocked?

Options

A. 0
B. `I_(1) // 2`
C. `I_(1)`
D. `2I_(1)`

Answer

To solve the problem, we need to analyze the situation involving three coherent light rays arriving at point P on a screen. The conditions given are that these rays produce an interference minimum of zero intensity when all three are present, and that if any two of the rays are blocked, the intensity at P is \( I_1 \).

### Step-by-Step Solution:

1. **Understanding the Interference Condition**:
- When three coherent light rays interfere at point P and produce a minimum of zero intensity, it implies that the rays are out of phase in such a way that they cancel each other out. This means that the resultant intensity from the three rays is zero.

2. **Intensity of Individual Rays**:
- Let the intensity of each individual ray be \( I_0 \). Since the rays are coherent and of equal intensity, we can denote the intensity of each ray as \( I_0 \).

3. **Condition with Two Rays Blocked**:
- When any two of the rays are blocked, the intensity at P is \( I_1 \). This indicates that the intensity from the remaining ray is \( I_1 \). Therefore, if one ray is left, it contributes its full intensity \( I_0 \) to the point P.

4. **Analyzing the Case with One Ray Blocked**:
- Now, if only one ray is blocked, two rays remain. Since we know that the original combination of three rays results in zero intensity, the two remaining rays must also interfere destructively.
- The two remaining rays will have a phase difference that causes them to partially cancel each other out.

5. **Calculating the Resultant Intensity**:
- When two rays are present, the resultant intensity can be calculated using the formula for the resultant intensity of two coherent sources:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- Here, \( I_1 \) and \( I_2 \) are the intensities of the two rays, and \( \phi \) is the phase difference between them. Since both rays have equal intensity, we can set \( I_1 = I_2 = I_0 \).
- Given that the resultant intensity when all three rays are present is zero, we can conclude that the phase difference between the two remaining rays must be such that they interfere destructively.

6. **Final Result**:
- When only one ray is blocked, the intensity at P will be \( 2I_0 \) (from the two remaining rays), but since they cancel each other out partially, the resultant intensity will be \( I_1 \) when one ray is blocked.
- Therefore, the intensity at P when only one ray is blocked is \( 2I_1 \).

### Conclusion:
The intensity of the light at point P, if only one of the rays is blocked, is \( 2I_1 \).