The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is

Options

A. `1:2:3`
B. `1:4:9`
C. `1:2/pi^2:3/pi^2`
D. `1:4/(9pi^2):4/(25pi^2)`

Answer

To solve the problem of finding the ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit, we can follow these steps:

### Step 1: Understand the Intensity Formula
The intensity of the maxima in a single slit diffraction pattern can be expressed as:
\[ I_n = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \]
where \( \beta = \frac{a \sin \theta}{\lambda} \) and \( I_0 \) is the maximum intensity at the central maximum.

### Step 2: Identify the Maxima
In the single slit diffraction pattern, the first secondary maximum occurs at positions where the path difference is an odd multiple of \( \frac{\lambda}{2} \). The positions of the maxima can be approximated using:
\[ \beta_n = \frac{(2n + 1) \pi}{2} \]
for \( n = 0, 1, 2, \ldots \)

### Step 3: Calculate Intensities for Consecutive Maxima
1. **Central Maximum (n=0)**:
\[ I_0 = I_0 \left( \frac{\sin(\frac{\pi}{2})}{\frac{\pi}{2}} \right)^2 = I_0 \]

2. **First Secondary Maximum (n=1)**:
\[ I_1 = I_0 \left( \frac{\sin(\frac{3\pi}{2})}{\frac{3\pi}{2}} \right)^2 = I_0 \left( \frac{-1}{\frac{3\pi}{2}} \right)^2 = \frac{4 I_0}{9\pi^2} \]

3. **Second Secondary Maximum (n=2)**:
\[ I_2 = I_0 \left( \frac{\sin(\frac{5\pi}{2})}{\frac{5\pi}{2}} \right)^2 = I_0 \left( \frac{1}{\frac{5\pi}{2}} \right)^2 = \frac{4 I_0}{25\pi^2} \]

### Step 4: Formulate the Ratios of Intensities
Now, we can write the ratios of the intensities of consecutive maxima:
- Ratio of Central Maximum to First Secondary Maximum:
\[ \frac{I_0}{I_1} = \frac{I_0}{\frac{4 I_0}{9\pi^2}} = \frac{9\pi^2}{4} \]

- Ratio of First to Second Secondary Maximum:
\[ \frac{I_1}{I_2} = \frac{\frac{4 I_0}{9\pi^2}}{\frac{4 I_0}{25\pi^2}} = \frac{25}{9} \]

### Step 5: Final Ratio of Intensities
Thus, the ratios of intensities of consecutive maxima can be summarized as:
- \( I_0 : I_1 : I_2 = I_0 : \frac{4 I_0}{9\pi^2} : \frac{4 I_0}{25\pi^2} \)

### Conclusion
The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is:
\[ 1 : \frac{4}{9\pi^2} : \frac{4}{25\pi^2} \]