For a parallel beam of monochromatic. Light of wavelength `'lamda'` diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the lightl. If 'D' is the distance of the screen from the slit, the width of the central maxima will be

Options

A. `(2Dlamda)/a`
B. `(Dlamda)/a`
C. `(Da)/(lamda)`
D. `(2Da)/(lamda)`

Answer

To find the width of the central maxima produced by a single slit diffraction, we can follow these steps:

### Step 1: Understand the setup
We have a single slit of width 'a' that is illuminated by a parallel beam of monochromatic light with wavelength 'λ'. The screen is placed at a distance 'D' from the slit.

### Step 2: Identify the angles
When light passes through the slit, it diffracts and creates a pattern on the screen. The angle θ at which the first minimum occurs can be approximated for small angles using the formula:
\[ \sin(\theta) \approx \theta \approx \frac{\lambda}{a} \]
This approximation holds true because the width 'a' of the slit is of the order of the wavelength 'λ'.

### Step 3: Relate the angle to the position on the screen
The position of the first minimum on the screen can be related to the angle θ and the distance D:
\[ y = D \cdot \tan(\theta) \approx D \cdot \theta \approx D \cdot \frac{\lambda}{a} \]
where 'y' is the distance from the center to the first minimum.

### Step 4: Calculate the width of the central maxima
The width of the central maxima (Δy) is defined as the distance between the first minima on either side of the central maximum. Since there is a minimum on both sides of the central maximum, we have:
\[ \Delta y = 2y = 2D \cdot \frac{\lambda}{a} \]

### Step 5: Final formula
Thus, the width of the central maxima is given by:
\[ \Delta y = \frac{2D\lambda}{a} \]

### Conclusion
The width of the central maxima for a single slit diffraction pattern is:
\[ \Delta y = \frac{2D\lambda}{a} \]

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