The line-width of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the line-width of a bright fringe in a Young's double slit experiment in terms of `lamda`, d and D where the symbols have their usual meanings

Explanation

To find the line-width of a bright fringe in a Young's double slit experiment, we will follow these steps:

### Step 1: Understanding the Concept of Intensity
In a Young's double slit experiment, the intensity of light at a point on the screen is given by the interference of light waves coming from the two slits. The maximum intensity occurs at the central maximum (bright fringe).

### Step 2: Maximum Intensity
Let the maximum intensity at the central bright fringe be denoted as \( I_0 \).

### Step 3: Condition for Half Maximum Intensity
The line-width of a bright fringe is defined as the distance between the two points on either side of the central maximum where the intensity falls to half of the maximum intensity. Thus, we need to find the positions where the intensity \( I \) is equal to \( \frac{I_0}{2} \).

### Step 4: Expression for Intensity
The intensity at a point on the screen in a double slit experiment can be expressed as:
\[ I = I_0 \cos^2 \left( \frac{\pi d \sin \theta}{\lambda} \right) \]
where \( d \) is the distance between the slits, \( \lambda \) is the wavelength of light, and \( \theta \) is the angle of the point from the central maximum.

### Step 5: Finding the Angle for Half Maximum
Setting \( I = \frac{I_0}{2} \):
\[ \frac{I_0}{2} = I_0 \cos^2 \left( \frac{\pi d \sin \theta}{\lambda} \right) \]
This simplifies to:
\[ \cos^2 \left( \frac{\pi d \sin \theta}{\lambda} \right) = \frac{1}{2} \]
Taking the square root gives:
\[ \cos \left( \frac{\pi d \sin \theta}{\lambda} \right) = \frac{1}{\sqrt{2}} \]
This implies:
\[ \frac{\pi d \sin \theta}{\lambda} = \frac{\pi}{4} \quad \text{or} \quad \frac{3\pi}{4} \]

### Step 6: Finding the Positions on the Screen
For small angles, \( \sin \theta \approx \tan \theta \approx \frac{y}{D} \) where \( y \) is the distance from the central maximum on the screen and \( D \) is the distance from the slits to the screen. Thus:
\[ \frac{\pi d \frac{y}{D}}{\lambda} = \frac{\pi}{4} \]
Solving for \( y \):
\[ y = \frac{\lambda D}{4d} \]
Similarly, for the other side:
\[ \frac{\pi d \frac{y'}{D}}{\lambda} = \frac{3\pi}{4} \]
Solving for \( y' \):
\[ y' = \frac{3\lambda D}{4d} \]

### Step 7: Calculating the Line-width
The line-width \( W \) is the distance between these two points:
\[ W = y' - y = \frac{3\lambda D}{4d} - \frac{\lambda D}{4d} = \frac{2\lambda D}{4d} = \frac{\lambda D}{2d} \]

### Final Answer
Thus, the line-width of a bright fringe in a Young's double slit experiment is:
\[ W = \frac{\lambda D}{2d} \]

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