In the Young's double slit experiment, the intensities at two points `P_(1)` and `P_(2)` on the screen are respectively `I_(1)` and `I_(2)` If `P_(1)` is located at the centre of a bright fringe and `P_(2)` is located at a distance equal to a quarter of fringe width from `P_(1)` then `I_(1)/I_(2)` is

Options

A. 2
B. `(1)/(2)`
C. 4
D. 16

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Setup
In the Young's double slit experiment, we have two points on the screen: \( P_1 \) at the center of a bright fringe and \( P_2 \) at a distance equal to a quarter of the fringe width from \( P_1 \).

### Step 2: Identify the Intensities
The intensity at point \( P_1 \) (which is at the center of a bright fringe) is the maximum intensity, denoted as \( I_1 \). The intensity at point \( P_2 \) can be expressed in terms of the phase difference due to its position.

### Step 3: Relate Path Difference to Phase Difference
The path difference \( \Delta x \) between the two slits for point \( P_2 \) is given by:
\[ \Delta x = \frac{d \cdot x}{D} \]
where \( d \) is the distance between the slits, \( x \) is the distance from the center of the fringe, and \( D \) is the distance from the slits to the screen.

### Step 4: Calculate the Phase Difference
The phase difference \( \Delta \phi \) is related to the path difference by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting for \( \Delta x \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \left( \frac{d \cdot x}{D} \right) \]
For point \( P_2 \), since it is at a distance of \( \frac{\lambda}{4} \) from \( P_1 \), we can express this as:
\[ \Delta x = \frac{\lambda}{4} \]

### Step 5: Substitute into the Intensity Formula
The intensity at any point in the interference pattern can be expressed as:
\[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
For point \( P_1 \):
\[ I_1 = I_0 \cos^2(0) = I_0 \]
For point \( P_2 \):
\[ I_2 = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
Substituting the phase difference for \( P_2 \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]
Thus,
\[ I_2 = I_0 \cos^2\left(\frac{\pi}{4}\right) = I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \]

### Step 6: Calculate the Ratio of Intensities
Now we can find the ratio of the intensities:
\[ \frac{I_1}{I_2} = \frac{I_0}{\frac{I_0}{2}} = 2 \]

### Final Answer
Thus, the ratio \( \frac{I_1}{I_2} \) is \( 2 \).

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