A parallel beam of light of wavelength `lambda` passes through a slit of width d. The transmitted light is collected on a screen D aways `(D gt gt d)`. Find the distance between the two second order minima.

Options

A. `(D lambda)/(d)`
B. `(D lambda)/(2d)`
C. `(2D lambda)/(d)`
D. `(4D lambda)/(d)`

Explanation

To find the distance between the two second order minima in a single slit diffraction pattern, we can follow these steps:

### Step 1: Understand the condition for minima
In single slit diffraction, the condition for minima is given by:
\[ d \sin \theta = n \lambda \]
where:
- \( d \) is the width of the slit,
- \( \theta \) is the angle of the minima,
- \( n \) is the order of the minima (for second order, \( n = 2 \)),
- \( \lambda \) is the wavelength of the light.

### Step 2: Substitute for second order minima
For the second order minima, we set \( n = 2 \):
\[ d \sin \theta = 2 \lambda \]

### Step 3: Use the small angle approximation
Since the distance \( D \) from the slit to the screen is much larger than the slit width \( d \) (i.e., \( D \gg d \)), we can use the small angle approximation:
\[ \sin \theta \approx \tan \theta \approx \theta \]
Thus, we can rewrite the equation as:
\[ d \theta = 2 \lambda \]

### Step 4: Solve for \( \theta \)
From the above equation, we can solve for \( \theta \):
\[ \theta = \frac{2 \lambda}{d} \]

### Step 5: Relate \( y \) to \( D \) and \( \theta \)
The position \( y \) of the minima on the screen can be related to \( D \) and \( \theta \) using:
\[ y = D \tan \theta \]
Since \( \tan \theta \approx \theta \) for small angles, we have:
\[ y \approx D \theta \]

### Step 6: Substitute \( \theta \) into the equation for \( y \)
Substituting \( \theta \) into the equation gives:
\[ y \approx D \left( \frac{2 \lambda}{d} \right) = \frac{2 \lambda D}{d} \]

### Step 7: Find the distance between the two second order minima
Since there are two second order minima (one on each side of the central maximum), the distance between them is:
\[ \text{Distance between two second order minima} = 2y = 2 \left( \frac{2 \lambda D}{d} \right) = \frac{4 \lambda D}{d} \]

### Final Result
The distance between the two second order minima is:
\[ \frac{4 \lambda D}{d} \]

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