In a Young's double-slit interference pattern, the intensity of the central fringe at `P =I_(0)`. When one slit width is reduced to half, the intensity at P will be

Options

A. `I_(0) // 2`
B. `I_(0) // 4`
C. `(I_(0))/(8)(3+2sqrt2)`
D. `(9)/(4)I_(0)`

Answer

To solve the problem, we need to analyze the situation step by step:

### Step 1: Understand the Initial Conditions
In a Young's double-slit experiment, the intensity of the central fringe (I₀) is determined by the contributions from both slits. Initially, we have two slits with equal width, and the intensity at the central fringe is given as I₀.

**Hint:** Remember that the intensity of light is proportional to the square of the amplitude of the wave.

### Step 2: Determine the Effect of Reducing One Slit Width
When one slit width is reduced to half, the amplitude of the wave coming from that slit will also be reduced. If the original amplitude from each slit is A, then the amplitude from the slit with reduced width becomes A/2.

**Hint:** The amplitude from each slit contributes to the total intensity at the central fringe.

### Step 3: Calculate the New Intensities
The intensity from the first slit (which remains unchanged) is proportional to A², and the intensity from the second slit (which is now half the amplitude) is proportional to (A/2)² = A²/4.

- Intensity from the first slit (I₁) = A²
- Intensity from the second slit (I₂) = A²/4

**Hint:** Use the relationship between intensity and amplitude to find the new intensity values.

### Step 4: Combine the Intensities
The total intensity at the central fringe (I) is the sum of the intensities from both slits:
\[ I = I₁ + I₂ = A² + \frac{A²}{4} = \frac{4A²}{4} + \frac{A²}{4} = \frac{5A²}{4} \]

**Hint:** Remember that when combining intensities, you simply add them together.

### Step 5: Relate the New Intensity to the Original Intensity
Since the original intensity I₀ was given as A² (when both slits had the same width), we can express the new intensity in terms of I₀:
\[ I = \frac{5}{4} I₀ \]

**Hint:** Substitute I₀ back into the equation to find the final answer.

### Final Answer
The intensity at point P after reducing one slit width to half is:
\[ I = \frac{5}{4} I₀ \]