The Miller indices of a plane having intercepts `2a, 2b,infty`are

Options

A. 110
B. 220
C. 100
D. 10

Explanation

To find the Miller indices of a plane with intercepts at \(2a\), \(2b\), and \(\infty\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Intercepts:**
The given intercepts are \(2a\), \(2b\), and \(\infty\).

2. **Write the Reciprocal of the Intercepts:**
The reciprocal of the intercepts gives us the vice indices:
- For \(2a\), the reciprocal is \(\frac{1}{2a}\).
- For \(2b\), the reciprocal is \(\frac{1}{2b}\).
- For \(\infty\), the reciprocal is \(\frac{1}{\infty} = 0\).

So, the vice indices are:
\[ \left(\frac{1}{2}, \frac{1}{2}, 0\right) \]

3. **Clear the Fractions:**
To convert the vice indices into Miller indices, we need to clear the fractions. We can do this by multiplying each vice index by the least common multiple (LCM) of the denominators. Here, the denominators are \(2\), \(2\), and \(1\), so the LCM is \(2\).

Multiplying each vice index by \(2\):
- \(\frac{1}{2} \times 2 = 1\)
- \(\frac{1}{2} \times 2 = 1\)
- \(0 \times 2 = 0\)

Thus, the Miller indices are:
\[ (1, 1, 0) \]

4. **Final Answer:**
The Miller indices of the plane with intercepts \(2a\), \(2b\), and \(\infty\) are \((1, 1, 0)\).